AN INTRODUCTION
TO INEQUALITIES
EDWIN BECKENBACH
RICHARD BELLMAN
The L. W. Singer Company
New Mathematical Library
AN INTRODUCTION
TO INEQUALITIES
by
Edwin Beckenbach
University of California, Los Angeles
Richard Bellman
The RAND Corporation
3
RAND O M
TH E
H O US E
L. W. SING E R
C O M PANY
Illustrations by Carl Bass
Fifth
Printing
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simultaneously in Toronto, Canada, by Random House of Canada, Limited.
School Edition published by The L. W. Singer Company
Library of Congress Catalog Card Number: 61-6228
Manufactured in the United States of A merica
To
Alan, Barbara, Edwin,
Eric, Kirstie, Lenann,
Suzann, and Tommy
NEW MATHEMATICAL LIBRARY
Other titles will be announced as ready
1. NUMBERS: RATIONAL AND IRRATIONAL by Ivan Niven
2. WHAT IS CALCULUS ABOUT? by W. W. Sawyer
3. INTRODUCTION TO INEQUALITIES by E. Beckenbach
and R. Bellman
4. GEOMETRIC INEQUALITIES by N.D.Kazarinoff
5. THE CONTEST PROBLEM BOOK, Problems from tne Annual
High School Contests of the Mathematical Association of
America, compiled and with solutions by Charles T. Salkind
6. THE LORE OF LARGE NUMBERS by P. J. Davis
7. USES OF INFINITY by Leo Zippin
8. GEOMETRIC TRANSFORMATIONS by I. M. Yaglon, trans­
lated from the Russian by Allen Shields
9. CONTINUED FRACTIONS by C. D. Olds
10. GRAPHS AND THEIR USES by Oystein Ore
11. HUNGARIAN PROBLEM BOOK, based on the Eotvos Com­
petitions, 1894-1905
12. HUNGARIAN PROBLEM BOOK, based on the Eotvos Com­
petitions, 1906-1928
13. EPISODES FROM THE EARLY HISTORY OF MATHE­
MATICS by Asgar Aaboe
14. GROUPS AND THEIR GRAPHS by I. Grossman and W.
Magnus
Note to the Reader
T
his book is one of a series written by professional mathematicians in
order to make some important mathematical ideas interesting and
understandable to a large audience of high school students and laymen.
Most of the volumes in the New Mathematical Library cover topics
not usually included in the high school curriculum; they vary in diffi­
culty, and, even within a single book, some parts require a greater degree
of concentration than others. Thus, while the reader needs little
technical knowledge to understand most of these books, he will have
to make an intellectual effort.
If the reader has so far encountered mathematics only in classroom
work, he should keep in mind that a book on mathematics cannot be
read quickly. Nor must he expect to understand all parts of the book
on first reading. He should feel free to skip complicated parts and
return to them later; often an argument will be clarified by a subsequent
remark. On the other hand, sections containing thoroughly familiar
material may be read very quickly.
The best way to learn mathematics is to
do
mathematics, and each
book includes problems, some of which may require considerable
thought. The reader is urged to acquire the habit of reading with paper
and pencil in hand; in this way mathematics will become increasingly
meaningful to him.
For the authors and editors this is a new venture. They wish to
acknowledge the generous help given them by the many high school
teachers and students who assisted in the preparation of these mono­
graphs. The editors are interested in reactions to the books in this
series and hope that readers will write to: Editorial Committee of the
NML series, in care of THE INSTITUTE OF MATHEMATICAL SCIENCES,
NEw YORK UNIVERSITY, New York 3, N.Y.
The Editors
CONTENTS
Note to the Reader
vii
3
Preface
5
Chapter
1
Fundamentals
Chapter
2
Tools
15
Chapter
3
Absolute Value
25
Chapter
4
The Classical Inequalities
47
Chapter
5
Maximization and Minimization Problems
79
Chapter
6
Properties of Distance
99
Symbols
1 13
Answers to Exercises
1 15
Index
131
AN
I N T R ODU C T I O N
T O
I N EQU ALI TI E S
Preface
Mathematics has been called the science of tautology; that is to
say, mathematicians have been accused of spending their time prov­
ing that things are equal to themselves. This statement (appropriately
by a philosopher) is rather inaccurate on two counts. In the first
place, mathematics, although the language of science, is not a science.
Rather, it is a creative art. Secondly, the fundamental results of
mathematics are often
inequalities rather than equalities.
In the pages that follow, we have presented three aspects of the
1, 2, and 3, we have the
4, we use the products of the
theory of inequalities. First, in Chapters
axiomatic aspect. Secondly, in Chapter
preceding chapters to derive the basic inequalities of analysis, results
that are used over and over by the practicing mathematician. In
Chapter
5, we show how to use these results to derive a number of
interesting and important maximum and minimum properties of the
elementary symmetric figures of geometry: the square, cube, equilat­
eral triangle, and so on. Finally, in Chapter 6, some properties of dis­
tance are studied and some unusual distance functions are exhibited.
There is thus something for many tastes, material that may be
read consecutively or separately. Some readers will want to under­
stand the axiomatic approach that is basic to higher mathematics.
3
4
P R E FAC E
They will enjoy the first three chapters. In addition, in Chapter
3
there are many illuminating graphs associated with inequalities.
Other readers will prefer for the moment to take these results for
granted and turn immediately to the more analytic results. They will
find Chapter
4 to their
taste. There will be some who are interested
in the many ways in which the elementary inequalities can be used
to solve problems that ordinarily are treated by means of calculus.
Chapter 5 is intended for these. Readers interested in generalizing
notions and results will enjoy the analysis of some strange non­
Euclidean distances described in Chapter
6.
Those whose appetites have been whetted by the material presented
here will want to read the classic work on the subject,
Inequalities,
by G. H. Hardy, J. E. Littlewood, and G. P6lya, Cambridge Univer­
1934 . A more recent work containing different
Inequalities, by E. F. Beckenbach and R. Bellman,
Ergebnisse der Mathematik, Julius Springer Verlag, Berlin, 1961.
sity Press, London,
types of results is
E. F. B.
R. B.
Santa Monica, California,
1960
C HAPT E R ON E
Fundamentals
1.1 The "Greater-than" Relationship
You will recall that the symbol "> " means "greater than" or "is
greater than. " Then you can readily answer the question: Is 3 > 2 ?
Of course it is.
But is
- 3 > - 2 ?Admittedly, - 3
is a "greater negative number"
than - 2, but this statement does not answer what is meant by the
question. If the real numbers (zero and the positive and negative
rational and irrational numbers) are represented geometrically in the
usual way by points on a horizontal number scale directed to the
right, as indicated in Fig.
-
I
5
I
-4
I
-3
I
-2
Figure
1. 1,
I
-1
then the numbers appear in order of
I
0
I I A real-number scale
increasing value from left to right. The point representing
to the right of the
- 2 > - 3. Similarly,
(l.l) 4 > -4 ,
3 > 2,
pears
point representing
O > -2,
- 3,
- 2 ap­
and accordingly
- 1 > -2,
1 > O.
Hence we have the following geometric rule for determining
inequality: Let a and b be any two real numbers represented by points
on a horizontal number scale directed to the right. Then a > b if and
5
6
A N I N T R O D UC T I O N T O I N E Q UA L I T I E S
only if the point representing a lies to the right ofthe point representing b.
You can say that - 3 > - 2 , or that - 300 > - 2 , but it is not
true according to the foregoing geometric rule.
In dealing with inequalities, it is often more fruitful and even nec­
essary to work algebraically instead of graphically. The geometric
rule given above has, in terms of the basic notion of positive number,
the following simple algebraic equivalent:
a and b be any two real
a - b is a positive number.
DEFI NITI ON. Let
and only if
Thus, if
a = -2
and
b = -3,
then
numbers. Then
a > b if
a - b = - 2 - ( - 3) = 1
is
positive. Hence - 2 > - 3 , as noted above in the geometric discus­
sion. You might check the inequalities in ( 1 . 1 ) by the present alge­
braic method of subtraction and verify each of the following inequa­
lities both by the geometric method and by the algebraic method:
77 > 3 ,
1 > - 9,
2 > o.
!
- > -40.
VI > 1 ,
1.2 The Sets of Positive Numbers, Negative Numbers, and Zero
You will note that in the preceding section we defined the inequality
a>b
in terms of positive numbers. The set P of positive numbers, and
similarly the set N of negative numbers, as well as the special set 0
having as its only member the number 0, play essential roles in the
study of inequalities. In fact, while of course we �hall freely use the
familiar algebraic (field) properties of the real number system, such
as the commutative, associative, and distributive laws, a basic thesis of
all order relationships in the real number system
-all algebraic inequalities-can be made to rest on two simple axioms
regarding the set P ofpositive numbers. These axioms will be presented
this entire tract is that
in the following section.
Symbolically, for "a is positive" we write "a e P," read in full as
"a is a member (or element) of the set P." Thus we have 5 e P, 0 e 0,
- 3 e N.
Let us look briefly at the foregoing sets, P,
members.
N,
and
The number zero, of course, is the unique member
it satisfies the equation
for any real number
a.
0,
and their
0 of the set 0;
FUNDA MENTA LS
7
Regarding the set N of negative numbers, it is important to dis­
tinguish the idea of the negative of a number from the idea of a nega­
tive number:
The negative of a number a is defined to be the number - a such
that
(a) + ( - a) = 0 .
a = - 3 then the negative of a is -( - 3) = 3 , since
( - 3) + (3) = 0 . Similarly, if a = 0 then - a = 0 since 0 + 0 = 0 .
A negative number is defined to be the negative of a positive num­
ber. Thus, you recognize 3, 1/2, 9/5, TT, VI as being members of the
set P of positive numbers. Then - 3, - 1/2, -9/5, -TT,- VI are
members of the set N of negative numbers.
Thus, if
We shall not attempt to define the basic notion of a positive num­
ber, but shall now proceed to characterize these numbers by means
of two basic axioms.
1.3 The Basic Inequality Axioms
The following simple propositions involving the set P of positive
numbers are stated without proof; accordingly, they are called
axioms. It is interesting to note that they are the only propositions
needed, along with the familiar algebraic structure of the real num­
ber system,t for the development of the entire theory of inequalities.
AxiOM I. Ifa is a real number, then one and only one ofthefollowing
statements is true: a is the unique member 0 of the set 0; a is a mem­
ber of the set P of positive numbers; - a is a member of the set P.
AXIOM II. If a and b are members of the set P of positive numbers,
then the sum a + b and the product ab are members of the set P .
The three alternatives listed in Axiom I relate an arbitrary real
number a and its negative - a as follows: If a is zero, then - a is
zero, as already noted; if a is positive, then - a is negative by the fore­
going definition of negative number; and if - a is positive, then
a = - ( - a) must be negative, again by the definition of negative
number. Thus a and - a are paired in the sets P, N, and 0 as indi­
cated in Table 1.
t But see the footnote on page 12.
8
A N INTR ODUCTION TO INEQU A LITIES
TABLE I. Pairings of Numbers and Their Negatives
Set
Number
a
p
N
0
-a
N
p
0
In the geometric representation (Fig . 1.1), the points representing
a and - a either coincide at the point representing 0 or lie on oppo­
site sides of that point.
1.4 Reformulation of Axiom I
Axiom I is concerned with the set P of positive numbers, and the
inequality a > b was defined in terms of the set P. Let us reformu­
late this axiom in terms of the inequality relationship.
If a and b are arbitrary real numbers, then their difference, a - b ,
is a real number; accordingly, Axiom I can be applied to a - b .
Thus either (a - b) e 0 (that is, a = b), or (a - b) e P (that is, a > b),
or - (a - b) = (b - a) e P (that is, b > a), and these three possibili­
ties are mutually exclusive. Hence, the following statement is a con­
sequence of Axiom 1:
AxiOM 1'. If a and b are real numbers, then one and only one of the
following relationships holds:
a = b,
a > b,
b > a.
In particular, Axiom I' asserts, in the special case b = 0 , that if a
is a real number then exactly one of the following alternatives holds:
a = 0 (that is, a e 0), or a > 0 (that is, a e P), or 0 > a (that is,
- a e P). Accordingly, Axiom I can be deduced from Axiom 1'.
If a statement Scan be deduced from-i.e., is a consequence of­
a statement T, we say "T implies S." We have just seen that Axiom I
implies Axiom I' and also that Axiom I' implies Axiom I. If each of
two statements implies the other, we say that they are equivalent.
Thus, Axioms I and I' are equivalent.
To illustrate Axioms I and 1', consider the numbers at = 3 ,
a2 = - 4 , bt = 0 , b2 = 3.
Illustrating Axiom I, you note that a1 e P, -a2 e P, bt e 0 , and
b 2 e P; you note also that a1 t 0 (read "a1 is not a member of the
set 0"), and - at t P, etc.
FUNDAMENTALS
9
Illustrating Axiom I', you have
a1 a1 a2a2 -
b1 =
b2 =
b1 =
b2 =
3 - 0 = 3,
3 - 3 = 0,
-4 - 0 = -4,
-4 - 3 = - 7 ,
a1 - b1 > 0 ,
a1 - b2 = 0,
b1 - a2> 0,
b2- a2> 0 ,
a1 > b1 ;
a 1 = b2;
bl >a2;
b2>a2.
You note, then, that in each of the four instances one and only
one of the three relationships given in Axiom I' holds. This illustra­
tion of Axiom I' will be continued in the following section, as addi­
tional inequality relationships are introduced.
1.5 AdditiQnal Inequality Relationships
In place of an inequality such as b> a, you might equally well
write a < b, read"a is less than b." The two inequalities are entirely
equivalent and neither is generally preferable to the other. In the
foregoing illustration of Axiom 1', the sign">" was used throughout
for the sake of consistency. But you might just as well have consid­
ered it preferable to be consistent in writing the a's before the b's in
all the relationships. Then you would have had
( 1 .2)
Likewise,
2 < 3.
- 2 < 0.
0 < 2,
-9 <
I,
-2 < - I ,
O<
I,
!
I<yl2, - 40 < - ·
The symbols">" and"<" represent strict inequalities.
Two other relationships considered in the study of inequalities are
the mixed inequalities a �band a s b , read"a is greater than or
equal to b" and"a is less than or equal to b," respectively. The first
of these, a �b, means that either a> b or a = b; for example,
3 �2 and also 2 �2 . The second, a s b, means that either a < b
or a = b; thus I s 2 and also 2 s 2 .
In ( 1 .2) it is stated that one of the three relationships listed in
Axiom I' holds in each instance. But the axiom itself asserts further
that only one of the relationships holds. Therefore, to make the illus­
tration of Axiom I' complete, you really should add the statements
( 1.3)
a1
i
b1 ,
a1 � b2,
a2 l b1,
a2 l b2,
read"a1 is neither less than nor equal to b1," etc.
AN INT RODUCTION TO INEQU A LITIES
You naturally feel that these negative statements in (1.3) are su­
perfluous, and indeed no one would claim that you should ordinarily
write them down to complete the information contained in (1.2). This
is due to the fact that the exclusiveness principle-the "one and only
one " aspect-of Axiom I or I' is taken for granted.
Because of the exclusiveness principle, the respective relationships
in (1.2) and (1.3) clearly are equivalent; that is, each implies the
other. Nevertheless, the negation of an inequality is often a very use­
ful concept.
If you tend to confuse the two symbols ">" and "<",you might
note that in a valid inequality, such as 3 > 2 or 2 < 3, the larger
(open) end of the symbol is toward the greater number while the
smaller (pointed) end is toward the lesser number.
10
1.6 Products Involving Negative Numbers
W hat sort of number is the product of a positive number and a
negative number? Or the product of two negative numbers? We can
use Axioms I and II and some of their consequences to determine
the answers to these questions.
If a e P and b e N , then - b e P according to Table 1, so that the
product a( - b) e P by Axiom II. Hence - [a( - b)] e N by the defini­
tion of a negative number; but - [a( - b)] = ab by the usual algebraic
rules for interchanging parentheses and minus signs:
- [a( - b)] = - [ - (ab)] = ab.
Therefore
abe N , and thus we have the following result:
THEOREM 1.1. The product ab of a positive number a and a negative
number b is a negative number.
Similarly, if a e N and be N , then - a e P and - be P according to
Table 1. Hence, by Axiom II, their product ( - a)( - b) e P. But by
the rules of algebra,( - a)( - b) = ab , and therefore abe P. Hence we
get this result:
THEOREM 1.2.
positive number.
The product ab of two negative numbers a and b is a
In particular, by this last result and Axiom II, the square of any
real number other than zero is a positive number. Of course, 02 = 0 .
FUNDAMENTA LS
II
Thus we obtain one of the simplest and most useful results in the
entire theory of inequalities:
THEOREM 1 . 3. Any real number a satisfies the inequality a2 �0.
The sign of equality holds if and only if a = 0 .
1.7 "Positive" and "Negative" Numbers
By this time, you may realize the strength of Axioms I and II. You
might be amused to learn that from them you could even determine
which of the nonzero real numbers belong to the set P of positive
numbers and which belong to the set N of negative numbers-as if
you didn't already know!
To see this, let us for the moment put "positive" and "negative"
in quotation marks to indicate information obtained from the axioms.
Let us start with the number a = 1 . Since a =F 0 , it follows from
Theorem 1 .3 that a2 > 0 . Thus a2 is "positive." But
a2 = 1 2 = 1 '
so that I is "positive."
Next, let us try a = 2 . Since we have now determined that 1 is
"positive," since 1 + 1 = 2 , and since by Axiom II the sum of two
"positive" numbers is "positive," it follows that 2 is "positive."
Now let a = t; then 2a = 1 . Th us the product of the "positive"
number 2 and the number a is the "positive" number 1 . But if a were
"negative," then the product of 2 and a would be "negative," by
Theorem 1.1 . Therefore a = tmust be "positive."
Thus the numbers 1 , 2, tare "positive" and hence, by Table 1, the
numbers - 1 , - 2, --!-are "negative."
Continuing, we can show that the integers 3 , 4, etc.; the fractions
J-, t, etc.; and the fractions f, 4, t. l• etc., are "positive," and accord­
ingly that - 3, - 4, -1. etc., are "negative." Thus, for any nonzero
rational number we can determine whether it is "positive" or
"negative."
Finally, the limiting processes that are used in defining irrational
numbers can be applied to determine, from our knowledge as to
which rational numbers are "positive" and which are "negative,"
whether a given irrational number is "positive" or "negative" in the
A N I N T R 0D U C T I
0 N T 0 I N EQ U A L l T I E S
completely ordered field of real numbers. t We shall not discuss irra­
tional numbers in any detail in this book; for an interesting treatment
of them, you should read Numbers: Rational and Irrational, by Ivan
Niven, also in this New Mathematical Library series.
12
Exercises
I. Make a sketch showing points representing the following numbers on a
horizontal number scale directed to the right:
3.
-1
.
- 1 .5. ., - 3 . 3 - .,
0.
•
V2. 2. - 2 . -3 .
Rewrite the numbers in increasing order, giving the result as a continued
inequality of the form a < b < c, etc.
2. Put a stroke throughe, thus f, if the
(a) - 3eN,
(b) 0e P,
(c) 5e0,
(d) v"1eN,
(e) (w - 3)eP,
statement is false:
( f ) a2e N,
(g) (a2 + 1 )e P,
(h) - 22eP,
(i) (a2 + l)e O,
(j)
-3e P .
3. Fill in each blank with P, N, or 0 so that a true statement results:
48 - 49 e
,
( f ) 72 - 4(2)(6)e
(a)
273
273
(b) 72 1 - 721 e
(g) 93(72 + t) - 93(72)e
837
838
- 25e --,
-23r- - ----n(h) 93(72 - t) - 93(72)e
(c) ---r
__
__
- 23 - ---r
- 23e
r(d) ----n(e)
-1 -
-2
I e
-2
,
---·
___
,
__
(i)
(
)
-- ,
2+3- 1 2+3
4 + 5 24 5 e
( j) ( -3)2 - 32e
__
,
__ ,
.
t Here ordered means that Axioms I and II are satisfied, and complete refers to the
basic property that, if a non-null set of real numbers has an upper bound, then it has
a least upper bound. For example, the set { l, 1 .4, 1 .4 1 , . . . } of rational approximations
to -../1 is bounded from above by 2, or by 1 .5, hence it has a least upper bound (which
we denote by y1) The corresponding point on the number scale (see page 5) divides
that scale into a left-hand and a nght-hand portion Since there is at least one "posi­
uve" rational number-e.g. I or 1 .4-in the left-hand portion, we say that -../1 is
"positive." In Sec. 1 .7 we have shown that the rational numbers can be ordered in only
one way, and we have stated that, similarly, the real numbers can be ordered in only
one way. The property of complete ordering, or its equivalent, is used in defining the
real numbers in terms of the rational numbers, and accordingly it is taken as a postu­
late for the real numbers rather than as an Axiom Ill for inequalities
F U N D A M E N TA L S
4. Fill in each blank with >. <.or = so that a true statement results:
49
48
(a)
(f ) 72
4(2)(6)'
273
273.
721
93(72),
(b) 721
(g) 93(72 + t)
837
838 '
- 25
(c) - 23
(h) 93(72 - t) -- 93(72).
---
--
JT -- )2"" ·
(d)
(e)
- 23
- 23
JT -- ]3 ·
-1
-2
-- - 2'
I
5 Label T for true or F for false:
(a) - 2 � -3
(b)
o::;o
(c)
O> - 1
(d)
(e)
�e )
2+3
+ 3
4 + 5--2 4 5'
(j)
( - 3)2
__
(f )
- 1::;2
( g)
t <t
(i)
-t < -t
-_
(i)
(h)
t<t
( j)
13
32 .
-! � -!
I - 22 < - 22
I<O
6. Wnte the negative of each of the following numbers:
- 2 3 -'" (3 - 7T)2 _a_' 0 Vb2 - 4ac
'
'
' b-c
'
·
7. Fill in the blanks to give affirmative relationships equivalent to the
stated negative relationships:
(a)
a { b, a
a
a
(b) a=f=b,
(c) a } b,
__
__
__
b,
b,
b,
(d) a i b,
(e) a l: b,
(f ) a�b,
a
a
a
__
__
__
b,
b,
b.
8. Show that each positive number pis greater than each negative number n.
9. For two real numbers a and b, what single conclusion can you draw if
you can show that a � b and that a::;b?
10. Prove by mathematical induction from Axiom II that if Ot. a2
an
are positive, then the sum a1 + a2 + · + an and the product a1a2 ···an
are positive. (See the comments in Sec. 2.6.)
• . . . •
• •
II. Use Axioms I and II to show that tis a "positive" number.
CHAPTER TW O
Tools
2.1 Introduction
In dealing with inequalities, the only basic assumptions that are
ultimately used are the two axioms discussed in Chapter l, along with
the real number system and its laws, such as the distributive law,
mathematical induction, etc. Still, there are several simple theorems
that derive from these axioms and that occur so frequently in the
development and application of the theory that they might almost
be called "tools of the trade."
These theorems,or operational rules, and their proofs are attrac­
tive and interesting on their own merits. Moreover, they furnish an
excellent illustration of the way mathematicians build up an elabo­
rate system of results from a few significant basic notions and as­
sumptions. The proofs are usually short, but nevertheless complete;
and in just a few places they call for the touch of ingenuity that makes
mathematics the fascinating subject that it is.
In the present chapter, some of these theorems are listed, illus­
trated, and proved. The letters a, b, c, etc., that occur in the state­
ment of the theorems will be understood to represent real numbers,
arbitrary except for explicitly given constraints.
For convenience, the theorems,or rules as we shall sometimes call
them, will be stated here only for the ">" case. In each instance,
15
16
A N I N TR 0 D U CTI
0N T0 I N EQ U ALITI ES
there is an equivalent "<" rule. Thus, to the ">" rule for transi­
tivity, "If a > b and b> c, then a > c," there corresponds the
equivalent "<" rule, "If a< b and b< c, then a < c ."
You can similarly construct the "<" rule that is the companion to
each of the other ">" rules given in this chapter, but watch out for
mathematical pitfalls! If a rule involves a positive multiplier, say
c > 0 , and requires that the difference of two arbitrary quantities be
positive, then the companion "<" rule still has c > 0 (or 0 < c if
you prefer), not c < 0 . Thus the companion rule to "If a > b and
c > 0 , then ac > be" is "If a < b and c > 0 , then ac < be."
The statement of the theorem at the beginning of each section is
divided into two paragraphs. The simpler first paragraph, dealing
with the strict ">" inequality sign, contains the heart of the result.
The second paragraph involves the mixed "� " inequality sign, and
in addition it sometimes includes the case of an arbitrary number n
of real values. Thus it deals with a more general and inclusive situa­
tion. The proof is usually given for the more inclusive general case
only, but it can readily be specialized to apply to the first paragraph.
The illustrations of the theorems given in this chapter will some­
times involve the stated ">" rule, and sometimes the implied "<"
rule.
2.2 Transitivity
If a> b and b> c, then a> c.
More generally, if a1 � a2 , a2 � aa, .. . , an -I� an , then a1 � an ,
with a1 = an if and only if all the a's are equal.
THEOREM 2.1.
Thus, if a consideration of your personal expenditures has led you
to observe that you spend more money on Saturday than on any
weekday, and that you spend at least as much on Sunday as on Sat­
urday, then you can conclude that you spend more money on Sun­
day than you do on any weekday.
Again, the solution of Exercise I in Chapter I is
-3 < -2 < - 1.5 < - 1 < 3 -
'IT
< 0 < 'IT
- 3 < V2 < 2 < 3 ;
this might be interpreted narrowly as meaning only that each of the
first nine members of the set is less than the immediately succeeding
number, thus: -3 < -2, -2 < - 1.5, etc. If so, it still implies, by
the foregoing transitivity rule, that each of these numbers is less than
TOOLS
any succeeding number; for example,
3 - 'TT < -J2.
17
-3 <
-
1 .5 , - 1 < 2 ,
PROOF. The transitivity rule could be proved by simple mathe­
matical induction (for which see the comments in Sec. 2. 6). But for
this first rule, we shall give a single direct proof involving four real
numbers.
Suppose, then, that a1 � a2 , a 2 � aa , aa � a4. By the algebraic
definition of inequality, each of the quantities a1 - a2 , a2 - aa,
a3 - a4 is either in the set P or in the set 0. Therefore the sum
(a1 - a2)+(a2 - aa)+(aa - a4) = a1 - a4
is either in P , by Axiom II, or in 0; and it is in 0 if and only
if a1 - a2 = 0 , a2 - aa = 0 , aa - a4 = 0 . Thus a1 � a4, and the
sign of equality holds if and only if a1 = a2 , a2 = aa, aa = a4.
The proof for the general case is left as an exercise.
2.3 Addition
THEOREM 2.2. If a> b and c > d, then a+ c > b+d. If
a > b, and c is any real number, then a+c> b+c.
More generally, if a1 � b1 , a2 d b 2 , . .. , an � bn , then
(2. 1)
a1+a2+.. +an � b1+ b 2+.. + bn .
·
·
The sign of equality holds in (2. 1 ) if and only if a1 = b1 , a2 = b2 , ... ,
an = bn .
Thus,if the inequalities I < y1. and 3 < ., are added, they yield
I +3 < Vi+., . This last inequality combined with - 1 = - 1
gives 3 < V2+., - I . And all five of these relationships added to­
gether produce 1 0 < 3 ( V2+.,) - 2 .
PRooF. As in the case of transitivity, an inductive proof is again
available. This time, however, a direct proof of the general case will
be given. Since by hypothesis each of the quantities a1 - b1 ,
a2 - b2 , . . . , an - bn e P or e 0, then by the generalization of Axiom
II given in Exercise 1 0 of Chapter I, the sum (a1 - b1)+(a2 - b2)+
+(an - bn) = (a1+a2+.. +an) - (b1 + b 2+ + bn) E P,
unless it e 0 with a1 - b1 = 0 , a2 - b2 = 0 , . . . , an - bn = 0 . Thus
"·
•
• • •
a1+a2+ +an � b1+ b2+
• • •
and equality holds if and only if a1
+ bn ,
• • •
= b1 , a2 = b2 , . .. , an = bn .
18
AN I N TR 0D UCTI0N T0 I N EQU ALITIES
2.4 Multiplication by a Number
THEOREM 2.3. If a > b and c > 0, then ac> be. If a > b and
c < 0 , then ac < be.
More generally, if a � b and c> 0 , then ac � be, with ac = be
if and only if a = b . If a � b and c < 0 , then ac �be, with
ac = be if and only if a = b .
Thus, multiplying the terms of an inequality by a positive number
leaves the sign of inequality unchanged, but multiplying by a nega­
tive number reverses it. In particular, for c = - I , if a � b then
- a � -b .
For example, if you multiply 3 >2 by 1 and -1, you obtain
3 >2 and -3 < -2 , respectively.
PRooF. It is given that a � b , so that a -be P or e 0. If ce P,
then it follows from Axiom II that c(a -b) = ca -cbe P or e 0 ;
that is, ca � cb .
But if ce N , then it follows from Theorem 1.1 that c(a -b) e N or
e 0 ; hence, - [c(a - b)] = cb - cae P or e 0 , so that cb � ca .
In either case, the sign of equality holds if and only if a = b.
2.5 Subtraction
THEOREM 2.4. If a> b and c > d, then a - d> b - c. If
a > b , and c is any real number, then a - c> b - c .
More generally, if a � b and c � d, then a - d� b - c, with
a - d = b - c if and only if a = b and c = d .
Note that dis subtracted from a, and c from b-not c from a, or
dfrom b.
Thus, by subtraction, the inequalities 7 > 6 and 5 >3 yield
7 - 3 > 6 - 5 , that is, 4 >1; but the inequality 7 - 5 > 6 -3
is false. Or, to illustrate the rule in terms of"<", the inequalities
-5 < 1 0 and -4 < -3 yield - 5 - ( -3) < 10 - ( -4); that is,
-2 < 14 .
PRooF. Applying the rule for multiplying an inequality by a nega­
tive number (Theorem 2.3) to c � d, we get -c � -d, that is,
- d � - c, where the sign of equality holds if and only if c = d.
Now applying the rule for adding inequalities t o a �b and
- d � - c, we obtain a+( - d)� b+( - c), or a - d� b -e,
with a -d = b - c if and only if a = b and c = d .
19
TOOLS
Exercises
I. Show that if a < b then a
2. Show that
and
< t(a
+
b) < b .
(a2 - b2)(c2 - d 2) � (ac - bd)2
(a2
+
b2)(c2 + d 2) :;::: (ac + bd)2
for all a, b, c, d, and that the signs of equality hold if and only if ad = be.
3. Show that
(a2 - b2)2 :;::: 4ab(a - b)2
for all a, b, and that the sign of equality holds if and only if a
= b.
4. Prove the general ">" transitivity rule by mathematical induction.
2.6 Multiplication
If a> b > 0 and c>d> 0, then ac> bd.
More generally, if a1 � b1 > 0 , a 2 � b2 > 0, . .. , an �bn> 0,
then
THEOREM 2.5.
(2.2)
The sign of equality holds in (2.2) if and only if a1
an = bn .
=
b1 , a2
=
b2 , . . . ,
Thus, from 2 >1 and 4 > 3 you obtain (2)( 4) >( 1 )(3) , or 8 >3 .
But note that -1 > -2 and - 3 > -4, yet ( -1)( - 3) < ( -2)( -4);
accordingly, the requirement that the numbers be positive cannot be
dropped.
PROOF. We shall give a proof by mathematical induction. The
standard procedure in constructing such a proof consists of the fol­
lowing steps. First, the statement to be proved for all positive
integers n is tested for the first one or two; then, under the assump­
tion that the statement is true for all integers up to and including a
certain one, say k - 1 , it is proved that the statement is true also
for the next integer,k. Since k can be any integer >1 (in particular,
let k - 1 = 1 or 2, for which the statement was verified), we may
conclude that the statement is indeed true for all positive integers.
For n = 1 , the conclusion a1 � b1 of Theorem 2.5 is simply a
repetition of the hypothesis. This observation is sufficient for the first
step in the proof by mathematical induction, but we shall give the
20
A N I N T R O D U C T I O N T O I N E Q U ALI T I E S
proof also for n = 2 ; that is, we shall show that if a1 � b1 > 0 and
a 2 � b2 > 0 , then a1a 2 � b1b2 .
The inequality
(2.3)
follows from the rule for the multiplication of an inequality by
a positive number; the sign of equality holds if and only if a1 = b1 .
The inequality
b1a2 � b1b2
(2.4)
follows from the same rule; the sign of equality holds if and only if
= b 2 . Now, the desired inequality
a2
(2.5)
follows from (2.3) and (2.4) by the transitivity rule (Theorem 2.1);
the sign of equality holds in (2.5) if and only if it holds in (2.3) and
(2.4), that is, if and only if a1 = b1 and a2 = b 2 .
We have demonstrated that the inequality (2.2) holds for n = I
and n = 2 .
Assume that the inequality (2.2) is true for n = 1 , 2 , . . . , k
I,
that is, for products of k - 1 numbers:
-
(2.6)
equality holds if and only if a1 = b1 , a2 = b2 , . . . , ak-1 = bk-1·
Then from the rule for multiplication by a positive number
(Theorem 2.3), when we multiply the inequality (2.6) by ak we obtain
(a1a2
(2.7)
• • •
ak-1)ak� (b1b2
• • •
bk-1)ak;
the sign of equality holds if and only if
a1a2
• • •
ak-1
=
b1b2
•• •
bk-1 .
Also, from the same rule, when we multiply the inequality
ak� bk
by
b1b2
•• •
bk-1 , we get
bk-1)bk;
the sign of equality holds if and only if ak = bk . It now follows
(2.8)
(b1b2
• • •
bk-1)ak� (b1b2
• • •
from (2.7) and (2.8), by the transitivity rule, that
ak-1ak� b1b2 bk-1 bk;
equality holds if and only if a1 = b1 , a2 = b2 , . . . , ak = bk.
a1a2
• ••
• ••
TOOLS
21
2.7 Division
THEOREM 2.6. If a>b>O and c>d>O, then ald>blc . In
particular, for a = b = I , if c>d> 0 , then 1 / d> 1 /c .
More generally, if a � b > 0 and c � d > 0 , then aid� blc,
with aid = blc if and only if a = b and c = d.
Note that a is divided by d, and b by c-not a by c, or b by d
Thus, by division, the inequalities 7 > 6 and 5 > 3 yield
713 >615 ; but the inequality 715 >613 is false.
The two given inequalities also yield 1 16 > 1 17 and 1 13 > 1 15 .
PROOF. We have
a
d
b
c
ac - bd
cd
The denominator cde P by Axiom II, since ce P and de P . Also,
since a� b and c� d, it follows from Theorem 2.5 that ac� bd;
hence the numerator ac - bde P or e 0, and actually ac - bde P
unless a = b and c = d . By Theorem 1 . 1 , the product of a nega­
tive number and a positive number is negative; but the product
cd ec ;d bd )
is equal to the nonnegative number
Hence
ac - bd, and cd is positive.
ac - bd
cd
is nonnegative. Thus
a = band c = d.
aid� blc, with aid = blc if and only
if
Exercises
l. From the inequality
show that
(Ja-�Yzo.
(1/a)
! (lib) S Vafi
for all positive a, b. Under what circumstances does the sign of equality
hold?
22
A N I N T R O DU CTION TO IN E Q U A L I T I E S
2. Show that the sum of a positive number and its reciprocal is at least 2;
that is, show that
a + l>
2
a for all positive values a. For what value of a does the sign of equality hold?
3. Show that
for all a, b, c.
a2 + b2 + c2 2 ab + be + ca
4. Show that
and
for all a, b.
5. Show that
a2b + b2c + c2a + ab2 + bc2 + ca2 2 6abc
for all nonnegative a, b, c.
6. Show that
(a2 b2) 2 2 (a - b)4
for all a, b satisfying ab 2 0 and that
(a2 - b2)2 � (a - b)4
for all a, b satisfying ab � 0 .
_
7. Show that
a3 + b3 2 a 2b + ab2
for all a, b satisfying a + b 2 0 .
8. For any of Exercises 3 through 7 that you have worked, determine under
what circumstances the signs of equality hold.
2.8 Powers and Roots
THEOREM 2. 7. If a > b > 0 if m and n are positive integers, and
if a11n and b11n denote positive nth roots, then
amln > b mln
and
b - mln > a - min
,
0
More generally, if a � b > 0, if m is a nonnegative integer and n
a positive integer, and if alln and b lln denote positive nth roots, then
(2.9)
amln � b mln
and
b - mln � a -min'
with a mln = b mln and b - mln
or (ii) m = 0 .
=
a-min if and only if either (i) a = b
23
TOO L S
For a = 9 and b = 4 , certain values of a mln, b m1n, b -mln, a-min
appear in Table 2. For each positive value of min you will note that
9mln > 4mln' while 4 - mln > 9 - mln
0
TABLE 2. Samples of Powers of Numbers
m
n
9 m1n
4 m1n
4 - mln
9 -mln
0
I
I
I
I
t
3
2
t
t
i
-h
t
!
..;.,
-n
-
I
t
2
9
4
27
8
8I
I6
PROOF. If m = 0 ' then a mln = b mln = b - mln = a -min = I ' so that
the sign of equality holds in (2.9) in this case.
If m =1= 0 , then a m � b m by the rule for multiplication of inequali­
ties (Theorem 2.5); the sign of equality holds if and only if a = b . If
it were true that alln < blln , then it would also be true that
(alln)n < (blln)n , or a < b; but, by hypothesis, a � b . Accordingly,
a vn � b lln . Therefore am1n � b mln , with am1n = b mln if and only
ifa = b.
For negative exponents, let
Then
amln =
a -min - !
-
b mln = d .
c,
c
'
Since we have just shown that
c
� d,
it follows from Theorem 2.6 that
that is,
the sign of equality holding if and only if c = d, that is, a = b .
The rule can be extended to positive and negative irrational
powers.
24
A N I N T R 0 D U C T1 0 N T 0 I N E Q U A L I T I E S
Exercises
1. Show that
(
)
a+b< a2 +b2 112
2 2
for all a, b. Under what circumstance does the sign of equality hold?
2. Show that if a, b, and e, dare positive (and e and dare rational), then
( ac - bc)( ad - bd) � 0
and
ac+d +bc+d � acbd + adbc
Under what circumstance do the signs of equality hold?
3. To what does the second inequality in Exercise 2 reduce in the case
e = d= I? in the case e = d= t?
4. For bd> 0,show that a/b::; e/d if and only if ad::; be , and that the
sign of equality holds in each place if and only if it does in the other
5. Show that if alb::; e/d, then
a+b<e+d
b - d '
and that the sign of equality holds if and only if ad= be.
6. Show that if alb ::; e/d, with a, b, e, d positive,then
_a_ < _ e _
a+ b-e + d'
and that the sign of equality holds if and only if ad= be.
7. Show that if alb$ e/d, with b and d positive, then
f!<a + e <f.
b-b + d-d'
and that the sign of equality holds if and only if ad= be.
8. Verify, by the test given in Exercise 4, that the four inequalities in the
conclusion of Exercises 5,6,and 7 are valid for the values a = 2 , b = 3 ,
e =5 ,d=6 .
9. Write the " <" inequality rule equivalent to the first paragraph of the
">"inequality rule for transitivity, addition, multiplication by a number,
subtraction, multiplication, division, and powers and roots.
CHA P T ER THR E E
Absolute Value
3.1 Introduction
In Chapter I , as you will recall, the inequality a > b was defined
in terms of the set P of positive numbers. You may also recall that
for the validity of several of the results of Chapter 2, such as Theo­
rem 2.5 concerning the multiplication of inequalities, it was necessary
to specify that certain of the numbers involved should be positive.
Again, in many instances the fractional powers of numbers that
appear in Theorem 2.7 would not even be real if the numbers them­
selves were negative; consider, for instance, a112 with a = -9 . Many
of the fundamental inequalities, which will be derived in Chapter 4,
involve just such fractional powers of numbers. It is natural, then ,
that we should often restrict our attention to positive numbers or to
nonnegative numbers (positive numbers and zero) in this study .
In applied problems involving inequalities, we often deal with
weights, volumes, etc., and with the magnitudes, or absolute values,
of certain mathematical objects such as real numbers, complex num­
bers, vectors. The magnitudes of all these are measured by nonnega­
tive numbers. Thus, even though you may choose to denote gains by
positive numbers and losses by negative numbers, a loss of $3 is still
a loss of greater magnitude than a loss of $2; the absolute value of
- 3 is greater than the absolute value of - 2.
25
26
A N I N T R O D U C T I O N TO I N E Q U A L I T I E S
In this chapter we shall define and study some properties of the
absolute value of real numbers, for application to inequalities in sub­
sequent chapters. We shall also exhibit the graphs of some interest­
ing but rather "off-beat" functions involving absolute values and
shall present some new ideas regarding them.
3.2 Definition
The absolute value of the real number a is denoted by I a 1 ; it can
be defined in a variety of equivalent ways. We shall consider several
of the defn
i itions here.
DEFI N ITION. The absolute value I a I of the real number a is defined
to be a if a is positive or zero, and to be - a if a is negative.
Thus,
121 = 2 , 101= 0 , and 1-21= - ( - 2) = 2 .
The principal disadvantage of the foregoing definition is that it is
unsuitable for algebraic manipulation. Thus (see Theorem 3.2 later
in this chapter) for all a, b it is true that
l a+b l �l a l+!b!,
as you can verify by considering separately the cases in which a and
b are both positive, one positive and the other negative, both nega­
tive, one zero and one positive, one zero and one negative, and both
zero. But it would be preferable to give a unifei d proof of such
a result by standard algebraic procedures; this will be done in Sec.
3.8, after a different but equivalent definition of absolute value in
terms of squares and square roots will have been given.
We could rephrase the above definition somewhat differently:
The absolute value I a 1 of the real number a is 0 if a e 0, and other­
wise I a I is the positive member of the set { a, -a} .
Thus, if a = 2 , then I a I is the positive member of { 2, - 2 } , i.e.,
2; if a= -2 , then I a! is the positive member of { 2
( 2) } i.e.,
2. But this characterization of I a I has the same algebraic disadvan­
tages as the preceding one.
-
,
-
-
,
3.3 Special Symbols
The next two characterizations of I a I depend on two useful spe­
cial symbols, max { } and { } +, which we shall now define.
27
A B S 0 L U T E VA L U E
For any set { Gt. a2 , . . . , an} of real numbers, the symbol
{a1, a2 , ... , an} denotes the greatest member of the set.
max
If there are only one or two members in the set,we still say
"greatest" in this connection; and if the greatest value is taken on by
more than one of the members,then any one of these is understood
to be the greatest.Thus,
max
{ 3. 7,0,-2,5}= 7 , max {4,4} = 4 , max { -3,-I}= -1 .
With some difficulty,arithmetic operations can be performed on
expressions involving the symbol max { } ; for example,
(max {4,-3}) (max {0, 5}) +max {-4, 4} -max { - 9, -8}
= 4_
2 max {1,4}
In particular,consider max
max
{a, -a}; if a= 2 ,then
{a, - a}= max {2,-2}= 2= I a I ;
if a= -3 ,then
max {a, -a}= max {-3,-(-3)}= 3
if
= I a I;
a= 0 ,then
max {a,-a}= max
{ 0,0} = 0 = I a I :
and so on. Thus,for all a,
( 3.1)
max
{a, - a } = I a I ,
so that ( 3.1) gives us another characterization of I a I ·
Let us turn now to the consideration of a second special symbol.
The symbol {a1. a2, .. . , an}+ denotes the greatest member of the set
{a1. a2 , ...,an} if there is at least one nonnegative member of the set;
but if all the members of the set are negative then the symbol denotes 0.
Thus
{ 3 . 7. 0,-2,5}+= 7'
{4,4}+
= 4,
{-3,
-1}+ = 0 .
As in the case of max { }. it is awkward but possible to deal
arithmetically with expressions involving the symbol { }+ ; for
example,
({4,
-3}+)({0, 5}+) +{-4,4}+-{- 9, -8}+
- 3.
2{1,4}+
28
A N I NTR0 D UCTI0N T0 IN EQUALITI ES
The symbols max { } and { } + are not equivalent, as the above
examples show; in fact, you can easily see from the definitions that
{ a1, a2, ... , an }+= max { 0, a1, a2 , ... , an }
= max
{
0,max
{ at, a2, .. . , an } [ .
Thus it follows that
where the sign of equality holds if and only if there is at least one
nonnegative member of the set { a1, a2 , . .. , an} .
Accordingly, since the special set { a, - a} does have a nonnega­
tive member for all a,
{ a, - a} + = max { a, - a } = I a I.
Thus, the equation
{ a, - a } + = I a I
might also be considered as a definition of I a I ·
Exercises
Determine the values of
(a ) max { -7 , -4, - I },
(b) max {3, .,, y'2} ,
(c) max {-7, 0, - I },
(d) max {0, 4, 1 },
(e) max {3, -3, 3},
2. For any set { a1. a2,
(f ) { - 7, - 4, - I )+,
(g) {3, 'Tr, yz}+'
(h) { -7, 0, - 1 }+,
(i) {0, 4, q+,
(j)
{3, -3, 3}+.
, On} of real values, the symbol min { Ot. a2, , an}
denotes the least member of the set { a1. a2,
, an}. and the symbol
{ a1. a2,
, an }- denotes the least member of the set { 0, a1, a2,
, an}.
Determine the values of
. •
• • •
• • •
.
.
(a)
(b)
(c)
(d)
(e)
•
•
min {-7, -4, - I },
min {3, .,, y'2} ,
min { -7, 0, - 1 },
min {0, 4, 1 },
min {3, - 3, 3},
(f)
(g)
(h)
(i)
{ - 7, - 4, - 1 }-.
{3, .,, vl2)-.
{ -7' 0, - 1 }-'
{0, 4, q-,
(j) {3, - 3, 3}-.
.
.
29
A B SO L U T E VA L U E
3 Determine the value of
(max { - I , - 2} )({ - I, - 2} +) -(min {I, 2} )({I, 2} - )
4. Show that
{
max max { a. b, c}, max {d. e}
} =max {
a,
b, c, d, e} .
5 Give an example showing that the inequality
max { a. h}
+
max { c, d}
�
max { a. b, c, d}
is not always valid.
6. Show that
{a, b} +
+
{c, d } + � { a, b, c, d } +.
7. Show that
{a1, a2, ... , an}+ � max {a�.a2 .... , an}
� min {a1, a2, ... , an }
� { a1.a2 . . . an}ls there any set for which the strict inequality sign holds in all three
places?
8. Show that if a= max {a, b, c}, then -a = min { - a, -b, - c}.
9. Show that { - a, -b}- = -{a, b)+.
10. Show that max {a1, a2, ... , an} = max a1, max {a2, a3, ... , an }
{
}
.
3.4 Graphical Considerations
A graphical representation can furnish a vivid and striking picture
of the behavior of a function, whether we are dealing with mean
daily temperatures, the fluctuations of the stock market, 1 xI, or
what-not; for one thing, a graph lets us see at a glance some of the
over-all properties of the function that otherwise might have been
obscure.
For example, the symbols max { } and { } + are made more
meaningful by a consideration of the graphs of
and
}
+
-�X-]_
4
4 '
which are shown in Figs. 3.1 and 3.2, respectively. In these figures,
30
A N I N T R O D U C T I O N TO I N E Q U A L I T I E S
the graphs of
1
--1
y =-x
2
2
and
3
7
-y = --x
4
4
are extended as dashed lines.
y
y
2
2
"
Graph of y
Figure 3.1.
= max
{ !2
x-
"
!,
-3 �X� 3
2
Graph of y
Figure 3 2.
{!2
=
x-
!
2'
-
-3 �X� 3
�x
4
-.:?.}+
4
'
Let us now construct the graph of the function defined by the
equation y = I x I· This graph provides us with a visual characteriza­
tion of absolute value. It will be sufficient for our purposes to restrict
our attention to the incomplete graph corresponding to the interval
-3�x�3 .
In the construction of this graph, it is helpful and interesting to
consider first the graph of y' = x, that is, the graph of the set
of ordered pairs of real numbers (x, y') such that y' = x, and also
the graph of y" = -x, as shown in Figs. 3.3 and 3 .4, respectively.
y
Y"
'
Figure 3.3.
'
Graph of y = x, . 3 � x � 3
-
Figure 3.4.
"
Graph of y
= -x,
-3 � x � 3
31
A B SO L U T E V A L U E
From these figures and the definition
lx l =max {x,
x} =max {Y',y"},
-
you can see at once that the graph ofy =I x I is simply the graph of
y =max {Y',y"},as illustrated in Fig. 3.5. Thus for each abscissa x,
the greater of the ordinatesy' andy" was chosen from Figs. 3.3 and
3.4 as the ordinatey in Fig. 3.5. For example,when x = -2 ,the
greater ordinate is y" = 2 ; when x = I ,the greater ordinate is
y' = I, etc.
Figure 3. 6 shows the graph ofy = - I x I ·
y
)'
:3
2
-I
-2
-:3
Figure 3 5.
Graph of y
=
I x 1. -3 �
x
�3
Graph of y
Figure 3 6
=
- 1 x 1. -3 �
x
�3
Looking at the four graphs in Figs. 3.3 through 3.6, you will note
that for no value of the abscissa x is any one of the corresponding
ordinates less than -1 x 1 or more than I x I · Explicitly, from Figs. 3.3,
3. 5, and 3. 6, you can read off the following result, which of course
you might have discovered and proved without considering the
graphs at all:
THEOREM 3.1.
For each real number a,
- l a l�a �l a l .
The first sign of equality holds if and only if a � 0 , and the second if
and only if a 2:: 0 .
Theorem 3.1 follows,for instance,from the fact that a = - I a I if
a e N or a e 0 , and a =I a I if a e P or a e 0 , and from the fact
(see Exercise 8 in Chapter I) that any positive number is greater
than any negative number.
32
AN I NTR0 DUCTI0N T0 IN EQU ALITI ES
Now, for exercises, let us consider the graphs of some more com­
plicated functions involving absolute values.
Consider first the graph of
y=
For x 2:: 0 ,
! <x+l x l ).
I x I = x , and hence
y = .! (x+x)= x;
2
but for
x < O , lx l = -x , so that
y = 2I (x - x) = 0 .
This graph, which is shown in Fig. 3.7, could easily have been pic­
tured from a consideration of Figs. 3.3 and 3.5 by taking the average
of the ordinates y', y for each abscissa x.
You might observe that the graph shown in Fig. 3. 7 is also the
graph of y= max { O, x}, as well as the graph of y = { xV. Thus,
{x}+ = max { 0, x } =
for all x.
Now let us look at the graph of
4 <x+l x l )
y= 2 1 x+ I I+l x l+l x - I I - 3
in the interval - 2 ::;; x ::;; 2 . For I ::;; x , the terms in the right-hand
member of (3.2) can be written thus:
(3.2)
2 1 x+I 1 = 2x+2 ,
so that for
lx l = x,
lx - I I = x - 1 ,
- 3 = -3 ,
1 ::;; x we have
y= 2x+2+x+x - I - 3= 4x - 2 .
For 0 < x < I , the first two members in the right-hand member of
(3.2) can be written as before, but
lx - I I = I - x,
not
lx - I I = x - 1 .
(Can you explain why?) Accordingly, for 0 < x < I ,
y = 2x+2+x+ I - x - 3 = 2x .
33
A B SO L U T E V A L U E
Similarly, for
- 1 :=;; x :=;;
0,
y = 2x + 2-x + I - x- 3 = 0 ;
and for x < -I ,
y = -2x
-
2 -x + I -x - 3 = -4x - 4 .
y
y
-I
-2
-3
-
I
2
Figure 3 8. Graph of
Figure 3.7.
Graph ofy = t<x + l x l ) , -3 � x � 3
y = 21x + l l + lxl + lx - 1 1 -3,
- 2 �X� 2
Thus eq. ( 3.2) is equivalent to a different linear equation in each of
the foregoing intervals. Plotting the corresponding linear segments in
the appropriate intervals,we obtain the continuous graph shown in
Fig. 3.8.
Exercises
I. From a consideration of Figs. 3.4, 3.5, 3.6, obtain a result for
ous to Theorem 3.I for a.
-a
2 For - 3 � x � 3 , sketch the graph s of
i
(a) y = (x-lx l ),
!
(b) y = < l x l - x) ,
3. Determine which of the graphs in Exercise 2 is also the graph of
(d) y =min {O, x} ,
(e) y =max {0, -x} ,
(f ) y =nun {0, -x} ,
(g) y ={x}- ,
(h) y ={ - x ) +,
(i) y ={ -x}- .
analog­
34
A N I N T R O D U C T I O N TO I N E Q U A L I T I E S
4. For 3 � x � 3 , sketch the graph of
(c) y = {x, -x - 2}+,
(d ) y = {x, -x - 2 } -.
(a) y = max {x, -x - 2 } ,
(b) y = min {x, -x - 2 } ,
5. For -3 � x � 3 , sketch the graph of
y = 2 1x - l l - lx1 + 2 1x + 1 1 - 5 .
6. A function f defined by an equatio n y = f(x) is said to be even if
/( - x) = f(x) for all x, and is s aid to be odd if f( -x) = -f(x) for all x.
Thus, the function defined by y = x2 is even since ( - x)2 = x2 , and the
function defined by y = x3 is odd since ( - x)3 = - x3; of course, some
functions are neither even nor odd. W hich of the functions illustrated in
Figs. 3.3 through 3.6 are even and which are odd?
3.5 The "Sign" Function
Another function closely related to I x I is illustrated in Fig. 3.9.
It is the function denoted by y = sgn x (read "sign of x," but not to
be confused with sin x), and defined by the equations
(3.3)
sgn x = + I
sgn x = 0
sgn x = -I
x > O,
X = 0,
x < O.
for
for
for
y
3
2
-3
- 2
-I
0
I
2
3
X
-2
-3
Figure 3 9 . Graph of y
=
sgn x , - 3 :.:::; X :.:::; 3
In the graph, the points corresponding to ( 0, I) and ( 0, -I) are en­
closed in unshaded, or open, circles to emphasize the fact that they
are not included in the graph, and the point corresponding to ( 0, 0)
is covered by a shaded circle to emphasize that it is included.
35
A BS 0 LU TE VA L U E
The functiony= sgn x is related toy= IxI through the notion of
slope, which is defined as follows :
Let L b e a nonvertical line in the coordinate plane, and let
P1 : (X t.JI) and Pz : (xz,yz) be distinct points on L; see Figs. 3. 1 0 and
3 . 1 1. In going from P1 to P2 , the vertical rise is the directed distance
yz - Y1 ,and the horizontal run is the directed distance xz-x1 =I= 0 .
Of course,either the rise or the run or both might be negative ; thus
in Fig. 3. II, the risey2 -y1 is negative (actually, then, a fall). The
ratio of the rise to the run, which has the same value for all pairs of
distinct points P1,P2 on L, is defined to be the slope m of L:
m
Figure 3.10
A line L having positive slope
z-Y1
=Y
.
Xz - X1
Figure 3. 1 1.
A hne L having negative slope
It should be easy for you to verify that the slopes of the linear
graphs ofy = x andy= - x are 1 and -I, respectively. See Figs.
3.3 and 3.4.
Now consider the slope of the graph of y= Ix I ,shown in Fig.
3. 5; at the same time, keep in mind the ordinate of the graph of
y= sgn x shown in Fig. 3.9.
For x > 0 ,the graph ofy= IxI coincides with the linear graph of
y = x , and it has slope m = I . Also, for x > 0 , the graph of
y= sgn x has ordinate y= I.
For x < 0 , the graph ofy = 1 xI coincides with the linear graph
of y= - x , and it has slope m = - I. Also, for x < 0 ,the graph of
y= sgn x has ordinatey= - I.
For x= 0 ,the slope of the graph ofy = Ix 1 is undefined, but you
can say that the right-hand slope at ( 0, 0) is I and the left-hand slope
at that point is - 1. The average of these slopes is 1[ 1 +(-I)]= 0 .
Also, for x= 0 , the graph of y= sgn x has ordinate y= 0 .
36
AN I NTRODUCTION TO I N EQUALITIES
Thus,y =
I x I andy = sgn x are related geometrically as follows:
For x =I= 0 , the value ofy = sgn x is equal to the value of the slope
of the graph y = I x I ; and for x = 0 , it is equal to the average of the
values of the right-hand and left-hand slopes.
It is interesting to note that although we have been analyzing two
relatively simple functions, y = 1 x I and y = sgn x , the graph of
y = 1 x 1 is somewhat peculiar in that it does not have a continuous
slope ; and the graph of y = sgn x is even more strange in that it is,
itself, discontinuous. We shall not here attempt to define continuity
and discontinuity, but the meaning should be intuitively clear in the
present instances.
The function defined byy = sgn x is related toy = 1 x 1 in a sec­
ond instructive way. A moment's reflection shows that, for each
real number a,
lal ;
accordingly, this equation gives us another characterization of the
absolute value I a I of a number a.
a sgn a
=
Exercises
I. For - 3 � x � 3 , draw lines passing through (0,0) and having slopes
(a) m = 0, (b) m = f , (c) m = - 1 .
2. Sketch the portions of the graphs of
(a) y = x ,
(d) y = x + 3 ,
(b) y = x + I ,
(c) y = x
(e) y = x - 1 ,
(f) J = X
that lie in the square - 1 � x � 1 , - 1 � y � 1 . It is
ble or extremely easy to do part (d) ; which? why?
3. For - 3 � x � 3 , sketch the graphs of
(b) y = x sgn (x
(a) y = (x + 1 ) sgn x ,
+ 2,
- 2,
either impossi­
+ 1) .
4. For - 3 < x < 3 , sketch an (x, y) graph such that, for each x, the value
ofy is the slope of the graph in Fig. 3.6; for any x where there is no defi­
nite slope, use the average of the right-hand slope and the left-hand slope.
3.6 Graphs of Inequalities
Before we leave the visually instructive subject of graphs, let us
investigate a few inequalities involving absolute values.
37
A B S O L U T E VA L U E
Consider,for example, the equati0n
l x l= 1
and the inequality
l x l :s:; l .
The equation has just two solutions, namely
x =-1 ;
and
x= 1
but the inequality has the entire interval
-I :s:; x :s:; I
as solution.
Again, the equation
l x- II= 2
has just the two solutions
X =-1
and
X= 3 ;
but the inequality
l x- 1 1 ::=:; 2
has, as solutions, each value x satisfying
-1 :=; x :=; 3 .
See Fig. 3.1 2.
-
4
-3
-2
-1
2
0
3
X
4
Figure 3. 1 2. Graph of the inequality 1 x
-
I I ::::; 2
+
To find the solution of the inequality
( 3. 4)
IX I+ly I ::::; I '
we consider values in each of the quadrants of the (x, y)-plane sepa­
rately. Thus in the first quadrant, where
x ;:::: 0
and
y ;:::: 0 ,
the inequality ( 3.4) is equivalent to
x +y :=; I.
38
AN I NTR 0DU CTI0N T0 IN EQ UALITI ES
We draw the portion of the line
x +y = l ,
i.e., of the line
y = I - X,
that lies in the first quadrant ; and since we seek the solution of
y ::;; I-X ,
our graph consists of those points in this quadrant that lie either on
or below the line.
The graph is shown in Fig. 3. 1 3, while the entire graph of the
inequality ( 3. 4) appears in Fig. 3. 1 4.
y
y
(0, - I )
Figure 3.14
Figure 3.13.
Graph of I x I + ly I :-::;; I
,
x � 0, y � 0
Graph of l x l + IY I :-::;; I
Figure 3. 1 3 might be considered also as showing the graph of the
solution of the set of inequalities
( 3. 5)
X +y ::;; I
x�O
y � 0.
In Fig. 3. 1 5 the shaded regions in (a), (b), and (c) comprise the solu­
tions, respectively, of the first, second, and third inequality in ( 3. 5);
the triply shaded region in (d) is the simultaneous solution of all the
inequalities ( 3. 5). Thus, though the set of equations
X +y = I
x =O
y =O
39
A BS 0 L U TE VAL U E
is satisfied by no pair of values (x, y),the set of inequalities ( 3.5) has
an entire area of solutions.
You will recall that, somewhat similarly, the equation Ix I = I has
only two solutions while the inequality IxI ::::; l has an entire interval
of solutions. Thus the set of all solutions of an inequality (or set of
inequalities) is often much "richer" than the set of all solutions of
the corresponding equation (or set of equations).
Exercises
I. In Fig. 3. 1 5(d), the plane is separated into seven differently shaded por­
tions. Each, together with its boundary, constitutes the solution of a set
of three inequalities; for example, one set is x � 0,y ;::: 0,x + y ;::: I .
y
0
(a)
X
( b)
y
0
X
(c)
Figure 3.15 Incomplete graph of (a) x + y ::::; I , (b) x � 0 , and (c) y � 0 ; tnply
shaded graph (d) of the intersection of (a), (b), and (c)
Give sets of inequalities for each of the areas.
2. Sketch the graphs of
(a) l x l - ly l ;::: I for - 2 � x � 2 .
( b)
l x l + 2 1y l � I .
40
AN I NTRODUCTION TO IN EQUALITI ES
3. Sketch the graph of the system
y �x+3
-2 � X � 2
y � 0.
4. Sketch the graph of the system
2x + y
x -y
X + 2y
5. Sketch the graph of the system
2x + y
x -y
X + 2y
� 5
� l
� 7.
� 5
� 1
� 7.
3.7 Algebraic Characterization
Our next, and last, characterization of 1 a 1 perhaps seems, at first
glance, to be the least desirable as a definition, for it appears to be
devious and unnatural; but it has the virtue of being algebraically
the most tractable, and accordingly it is the one we shall most often
use.
To this end, consider the following: If a= -2 ,then
a2 = ( - 2)2= 4 ,
if
a= 0 , then
a2 = 02 = 0 ,
yl4= 2 = I a I,
so that
y;i2= l a l ;
yo= 0= Ia I,
so that
P= l a l ;
so that
P= l a l .
a= 2 , then
a 2 = 22 = 4 ,
and if
Similarly for all real a,
P= l a l ,
and this is our algebraic characterization of Ia I·
Actually the characterization Ia I= ...jQ2 expresses the special case
b= 0 of the Pythagorean relationship,
c= ya2 +b2 ,
between the lengths a, b of the sides and the length c of the hypote­
nuse of a right triangle. Thus the absolute value of the real number
a can be interpreted as the length, or magnitude, of the line segment
from the origin to the point representing a on the number scale
A B S O L U T E VA L U E
41
(Fig. 1 . 1). The reader who continues to study mathematics will soon
learn that this last characterization of absolute value (and the accom­
panying geometric concept of length) can be appropriately general­
ized so that absolute value may be defined also for other mathematical
objects, e.g., vectors and complex numbers.
Two points to note in the foregoing algebraic expression of Ia I
are, first, that a2 is nonnegative, so that its square roots are real, and
secondly,that by definition the symbol y means the nonnegative
square root.To be sure,(-+- 2)2 = 4 ,so that 4 has two square roots­
namely -+- 2 ; but in algebraic manipulations the symbol y'4 denotes
only 2, not -2. For example,consider the equations
and
5 + V'f= 7 ,
5 - vf4 = 3 ,
5 - Vlf = 7 ,
5 + V'f = 3 .
The first two of these equations we consider to be valid, and the sec­
ond two invalid,but only because the symbol y always means the
nonnegative square root. It is for this reason,also,that -+- appears
before the symbol y in the familiar formula for the solutions of
the quadratic equation
namely
ax 2 + bx + c = 0 ,
2
x - - b -+- yb - 4ac
2a
_
·
Exercises
I. You can see from the Pythagorean relationship that, for the equation
x2 + y2 = r2 ,
the solution set (locus) of points (x,y) constitutes the circle with center at
the origin and radius r. Determine the solution set of values (x, y) for the
inequality
x2 + y2 � 25 .
2. The absolute value of the complex number x + iy is defined by
I x + iy I = yx2 + y2 .
Representing x + iy by the point having coordinates (x, y) in the plane,
determine the solution set for
I � l x + iy l � 2 .
3. Determine the locus of points x + iy for which
l x + iy + I I = l x + iy - 1 1 .
AN l NTR 0D U CTl 0N T0 l N EQ U AL lTl ES
42
3.8 The "Triangle" Inequality
The inequality ( 3.6) in Theorem 3.2 below, to which we have
referred earlier in this chapter,is often called the "triangle" inequal­
ity for geometric reasons that will be discussed further in Chapter 4.
Here is a complete statement concerning the inequality:
THEOREM 3.2. For all real numbers a and b,
lal
+ lbl � la + bl.
The sign of equality holds if and only if ab � 0 , that is, if and only if
a and b are either both � 0 or both s O.
( 3. 6)
For example,if
while
But if
a = 5 and b = - 2 , then
Ia + bI = 1 5 +
< - 2) I
= 3.
Ia I + lbl = 1 5 1 +1 -2 1 = 7.
a = - 5 and b = - 2 , then
I a + b l = 1 ( - 5) + ( - 2) 1 = 7
and
IaI + lbl = 1 -51 + 1 -21 = 7 .
You can grasp the validity of the inequality ( 3. 6) intuitively by
noting that if a and b are of opposite sign, then on the right-hand
side of the inequality the numbers a and b will "work against each
other" and emerge from the encounter with diminished joint magni­
tude,whereas on the left-hand side they are forced to "pull together"
positively from the start.
On the same basis,you can understand the inequality
( 3.7)
la
- b l � i l a l-l b l l .
in which the sign of equality holds again if and only if ab � 0 (i.e.,
a and b are both � 0 or both s 0). Here, on the right-hand side of
the inequality, the numbers a and b are always forced to "settle their
difference," whereas on the left-hand side,if a and b are of opposite
sign, then they will pull together to increase their joint magnitude.
Inequalities ( 3. 6) and ( 3.7) can be proved by using the algebraic
definition I a I = ....ji.i2 Thus inequality ( 3.6) can be restated equiva­
lently as
( 3.8)
43
A B SO L U T E VA L U E
Now, ( 3.8) is equivalent to
( yaz + Vfll)2 2 (a + b) 2 ;
( 3.9)
that is, the validity of either ( 3.8) or ( 3.9) implies that of the other.
Thus ( 3.9) follows from ( 3.8) by squaring (see Theorem 2.5), and ( 3.8)
follows from ( 3.9) by taking nonnegative square roots (see Theo­
rem 2.7).
Next,( 3.9) can be rewritten as
a2
+
2 -.jii252 +
b2 2 a2
+
2ab +
b2 .
Accordingly, ( 3.9) is equivalent to
( 3.1 0)
-.J(i252 2 ab
by the inequality rules for addition,subtraction, and multiplication
by a positive number. But
-.Jii252 = yl(ti7))2 = I ab I ,
so that ( 3.1 0) is equivalent to
( 3.1 1)
l ab l 2 ab.
Thus ( 3.6) is equivalent to ( 3. 1 1).
Now, ( 3. 1 1) is valid by Theorem 3. 1, which states that any real
number is less than or equal to its absolute value. The sign of equality
in ( 3. 1 1 ) holds if and only if ab 2 0 . Therefore the inequality ( 3. 6),
which is equivalent to ( 3.1 1),is also valid, and the sign of equality
holds if and only if ab 2 0 .
The inequahty ( 3.7) can be proved similarly after it has been written
in the equivalent form
y(a - b)2 2
J< Vfi2 - y'b2)2 .
However,it is interesting to note that ( 3.7) can also be derived
directly from ( 3.6). Thus, substitution of the real number a - b for
the arbitrary real number a in ( 3. 6) yields
or
whence
Ia - bl + lbl 2 1a - b + bl,
I a - b l + lb l 2 1 a l ,
( 3.1 2)
by Theorem 2. 4, the subtraction rule. Similarly,substitution of
b-a
44
AN I NTRO DUCTION TO I N EQUALITI ES
for b in ( 3. 6) yields
IaI + lb - al z la + b - al,
or
lal + la - bl z lbl,
whence
l a - b l z lbl - l a l .
( 3.1 3)
Since
j l a l - l b l j = max { ( l a l - l b l ) , - ( l a l - l b l )}
= max { < I a I - I b I ) , < I b I - I a I ) } ,
it follows from ( 3.1 2) and ( 3.1 3) that
la - bl z l lal - lbl
Hence ( 3. 7) is valid,the sign of equality holding if and only if it holds
either in ( 3.1 2) or in ( 3.1 3).
Since a - b was substituted for a in deriving ( 3.1 2), the sign of
equality holds in ( 3.1 2) if and only if (a - b)b 2 0 , i.e.,ab 2 b2 . The
latter inequality is valid if and only if ab 2 0 and I a I 2 I b I · Simi­
larly,the sign of equality holds in ( 3. 1 3) if and only if a(b - a) 2 0 ,
that is,ab 2 a2 • This is valid if and only if ab 2 0 and l b l z l a l .
Since at least one of the inequalities I a 1 2 I b I and I b I 2 I a I holds,
it follows that the sign of equality holds in ( 3.7) if and only if ab 2 0 .
It might be noted that since b represents an arbitrary real number­
positive,zero,or negative-the inequalities ( 3.6) and ( 3.7) still hold
if b is replaced by -b. We thus obtain
( 3.1 4 )
lal + lbl z la - bl
from ( 3. 6), and
( 3.1 5 )
from ( 3.7), with the signs of equality holding if and only if
a( - b) 2 0 , that is,ab � 0 . Together,inequalities ( 3.6), ( 3.7), ( 3. 1 4),
and ( 3.1 5) can be written as
(3. 16)
I a I + l b l z l a -+- b i z l l a l - l b l l ·
45
A B S O L U T E VA L U E
Exercises
I. From the algebraic characterization of absolute value, show for all real
a, b that
(a) 1 -a l = l a l ,
(b) l ab l = l a l · l b l ,
and if b =F 0 that
(c)
l,
a _ lal
h
fbi '
2 Deterrrune whether it is the sign < or the sign = that holds in the mixed
inequality I a - b l � l l a l - l b l l , i f
(d) a = - 9 , b = - 1 0 ;
(a) a = 'TT ,
b = 2'TT ;
(e) a = 9 ,
b = - 10 .
(b) a = -'TT , b = y'2;
(c) a = 2 ,
b = O;
3. Determine whether it is the sign > or the sign = that holds in the mixed
inequality I a I + l b l � I a + b l , if
(d) a = 0 , b = - 2 ;
b = -2 ;
(a) a = 3 ,
(e) a = O , b = 0 .
(b) a = - 3 , b = - 2 ;
(c) a = 3 ,
b = 2;
4. Repeat Exercise 2 for the inequality I a + b I � I I a I - I b I I .
5. Repeat Exercise 3 for the inequality I a I + I b I � I a - b I ·
6. Show that the inequality I a - b I �
equality I ab I � ab .
I I aI - I bI I
is equivalent to the in­
7. Show that if ab � 0 then ab � min { a2, b2 } .
8. Show that each o f the other characteristic properties o f I a I , given in this
chapter, follows from I a I = Vfi2 .
C HAPT ER FOUR
The C lassical Inequalities
4.1 Introduction
Now that we have forged our basic tools,we shall demonstrate more
of the magic of mathematics. As an artist evokes, out of a few lines
on a canvas, scenes of great beauty, and as a musician conjures up
exquisite melodies from combinations of a few notes,so the mathe­
matician with a few penetrating logical steps portrays results of
simple elegance. Often, like the product of the magician's wand, these
results seem quite mysterious,despite their simplicity, until their
origin is perceived.
In this chapter, we shall employ the basic results derived in the
previous chapters to obtain some of the most famous inequalities in
the field of mathematical analysis. These inequalities are the every­
day working tools of the specialist in this branch of mathematics.
In Chapter 5, we shall then show how these new relationships may
be used to solve a number of interesting problems that, at first sight,
seem far removed from algebra and inequalities. The applications are
continued in Chapter 6, where we discuss and extend the notion of
distance.
This,indeed, is one of the fascinations of mathematics-that simple
ideas applied one after the other,in the proper order,yield results
that never could have been envisaged at the outset.
47
48
A N I NT R O DU CTION TO I N E Q U A L I T I E S
4.2 The Inequality of the Arithmetic and Geometric Means
(a) Mathematical Experimentation. Given two nonnegative num­
bers, say I and 2, let us obtain their "mean" in the following two ways:
the arithmetic mean (or half their sum), usually called "average,"
I+ 2
2
=
1.5 ,
and the geometric mean (or square root of their product )
yr:2
=
1.4 1 · · ·
0
Observe that 1. 5 > 1. 4 1
Similarly, if we start with the numbers
3 and 9, for the arithmetic mean we obtain the value !( 3 + 9) = 6 ,
and for the geometric mean we get the value VJ.1 = 5. 19 · · · .Note
that 6 > 5.19
Continuing with various pairs of nonnegative
numbers chosen at random, say I I and 1 3, t and t . and so on, we
observe in each case that the arithmetic mean is greater than the
geometric mean.
Can we safely generalize this discovery and reach some conclusions?
Our mathematical nose begins to twitch as we scent a theorem.
Maybe the result is true for all pairs of nonnegative numbers ! In
other words, we conjecture that the arithmetic mean of two nonnega­
tive numbers is always at least as great as their geometric mean. We
shall express this conjecture in terms of algebraic symbols and we
shall see in subsection (b) that our conjecture is true. We may there­
fore state it as
• • • .
• • • .
THEOREM 4.1. For any nonnegative numbers a and b,
( 4.1)
a + b > -y!{r
uv .
2
_
The sign of equality holds if and only if a = b .
Note that if one of the two numbers were positive and the other
negative, then ( 4.1) would be meaningless since its right-hand side
would be imaginary.t If both numbers were negative, then the left­
hand side of ( 4. 1) would be negative and the right-hand side positive,
so that the theorem would not be valid.
t The notion of inequality is not directly applicable to imaginary numbers, but only
to their absolute values
T H E CLASSI CAL I NEQ UALITI ES
49
The type of experimentation that led us to Theorem 4.1 repre­
sents the sort of trial-and-error methods often used by mathemati­
cians on the trail of theorems. Formerly, it was quite laborious work.
Nowadays,with the modern digital computer to aid mathematical
experimentation,we can test thousands and millions of cases in a few
hours. In this way, we obtain valuable clues to general mathematical
truths.
Exercises
l . Determine the geometric mean and the arithmetic mean for the following
pairs of numbers:
(a) 2, 8,
(c) 4, 9,
(d) 0, 20.
(b) 3, 1 2,
2. If p is nonnegative, determine the geometric mean and the arithmetic
mean of the following pairs of numbers:
(c) 2, 2p2.
(a) p, 9p,
(b) 0, p,
(b) Proof of the Arithmetic-mean-Geometric-mean Inequality for
Two Numbers. Since square roots are a bit bothersome,let us elimi­
nate them by writing
a = c2 ,
( 4.2)
This is permissible because a and b were assumed,in Theorem 4. 1,
to be nonnegative. The relationship ( 4.1) that we wish to prove for
arbitrary nonnegative a, b then becomes
c2+d 2 ?:. cd
2
( 4. 3)
for arbitrary real c, d. Now ( 4.3) is true if and only if
c2+d 2
- cd ?:. 0 ,
2
( 4.4)
which is equivalent to
( 4.5)
c2 + d 2 - 2cd ?:. 0
as a result of our elementary rules for dealing with inequalities.
We now recognize a familiar friend,namely
( 4.6)
c2+d 2 - 2cd = (c - d) 2 ,
so that ( 4.5) is equivalent to
( 4.7)
(c - d)2 ?:. 0 .
50
AN I N T R O D U CTION TO IN EQUALITIES
Since, by Theorem 1.3, the square of any real number is nonnega­
tive, we see that ( 4.7) is indeed true. Thus ( 4.5) is a valid inequality,
and hence ( 4.4), ( 4.3), and ( 4.1) are also valid. The sign of equality
holds in ( 4. 7), and therefore in ( 4. 1), if and only if c d = 0, that is,
c = d, or, equivalently, if and only if a = b .
Note that, while the inequality ( 4.1) of Theorem 4.1 applies only
to nonnegative numbers a, b, the foregoing proof shows that in­
equality ( 4.3) is valid for all real numbers c, d, the sign of equality
holding if and only if c = d . You will observe that the results of
Sees. 4. 4 and 4.6 also are valid for all real numbers, not merely non­
negative numbers; this fact increases the geometric significance of
those results.
-
(c) A Geometric Proof Let us now show that Theorem 4.1 can also
be obtained geometrically by means of a simple comparison of areas.
Consider the graph of y = x , as shown in Fig. 4.1. Let S and T be
points on the line y = x , with coordinates (c, c) and (d, d), respec­
tively, and consider the points P : (c, 0) , Q : (0, d) , and R : (c, d) , as
shown. Since OP is of length c, PS has the same length c. Then the
area of the triangle OPS is c2 / 2 , i.e., one half of the base times the
altitude. Similarly, the area of the right triangle OQT is d 2 !2.
c
Figure 4. 1 Geometric proof of the inequality
c2
i d 2 :::>: cd
Now examine the rectangle OPRQ. Its area is completely covered
by the triangles OPS and OQT, so that
( 4.8)
area (OPS ) +area (OQT) 2 area (OPRQ) .
Since the area of OPRQ is cd, length times width, we may write ( 4.8)
THE CLASSICAL INEQUALITIES
in algebraic symbols as
51
c2 i d 2 2 cd .
(4.9)
Now the inequality (4.9) is identical with the inequality (4.3), so that
our geometric proof is complete.
Furthermore, we see that there is equality only when the triangle
TRS has area zero,that is, only when S and T coincide, so that
c=
d.
(d) A Geometric Generalization. A little thought will show that the
foregoing arguments remain valid even in cases where the curve OTS
is not a straight line. Consider the diagram shown in Fig. 4.2. It is
still true that
(4. 10)
area (OPS) +area (OQT) 2 area (OPRQ) .
Figure 4.2. A more general geometnc inequality
When you have studied calculus and have learned how to evaluate
the area underneath simple curves, such as y = xa , for arbitrary
positive a, you will find that this process will yield a number of in­
teresting inequalities in a very simple fashion. In later sections of this
chapter, we shall obtain some of these inequalities in a different way.
Exercises
l . Let a and b be the lengths of a pair of adjacent segments on a line, and
draw a semicircle with the two segments together as its diameter, as in
1- ..,.
0
0
b
Figure 4.3
----'•10-ll
Fig. 4.3. Show that the radius r of the circle is the arithmetic mean of a
and b, and that the perpendicular distance h is their geometric mean.
52
AN I NTR0 DU CTI0N T0 IN EQ U ALITI ES
2. An average that occurs quite naturally in optics and in the study of elec­
trical networks is the harmonic mean. For two given positive quantities
a and b, the quantity c determined by means of the relationship
� =l+l
c a b
is called the harmonic mean. Solving this equation for c, we obtain
2
_ 2ab
c1 /a + 1 /b - a + b ·
Show that the harmonic mean is less than or equal to the arithmetic mean,
and also less than or equal to the geometric mean, with equality if and
only if a = b ; i.e., show that
a + b>
> 2ab .
2 - Vfi6 - a + h
3. Determine the harmonic, geometric, and arithmetic means of the pairs
(b) 3, 1 2;
(c) 4, 9;
(d) 5,7 ;
(e) 6, 6.
(a) 2, 8;
4. The relationship between distance d, rate r, and time t is d = rt . Show
that if in traveling from one town to another you go half the distance at
rate r�> and half at rate r2 , then your average rate is the harmonic mean
of r1 and r2, but that if you travel half the time at rate r1 and half the time at
rate r2, then your average rate is their arithmetic mean. If r1 =F r2 , which
method would get you there sooner?
5. Use the result of Theorem 4.1 to solve Exercise 2 on page 22.
(e) The Arithmetic-mean-Geometric-mean Inequality for Three
Numbers. Let us now perform some further experimentation. Taking
three nonnegative numbers, say 1 , 2, and 4, let us compute their
arithmetic mean-the simple average-as before:
1 +2+4
3
=
2. 33 • • • .
Let us also compute their geometric mean, i.e., the cube root of their
product:
�1 ·2·4
=
2.
We observe that the arithmetic mean of these three numbers is
greater than their geometric mean. Performing a number of such ex­
periments with triplets of nonnegative numbers, we constantly observe
the same result. We begin to suspect that we have found another
theorem. Can it be true that there is an extension of Theorem 4. 1, a
result asserting that the arithmetic mean of three nonnegative quan-
53
T H E C LASSICAL I N EQ U A L I T I ES
tities is at least as great as their geometric mean?
We wish to prove
THEOREM
4.2. For any three nonnegative numbers a, b, c,
(4. 1 1 )
The sign of equality holds if and only if a = b = c .
To remove the cube roots, let us set
c = z3 .
a = x3 ,
(4. 12)
a, b, and c in (4. 1 1 ),
3
x + y3 + z3
:::::: xyz ,
3
Substituting these values for
(4. 1 3)
we obtain
which is equivalent to
x3 + y3 + z3 - 3xyz :::::: 0 .
(4. 14)
We shall prove Theorem 4.2 by proving the validity of
arbitrary nonnegative x, y, z.
(4. 14) for
Once again we have an expression that can be factored. Its factori­
zation is not as common as the one used before, but is still a quite
useful one. We assert that
(4. 15)
x3 + y3 + z3 - 3xyz
= (x + y + z)(x2 + yz + z2 - xy - xz - yz) ,
a result that can be verified by multiplication.
Since
x +y + z
is nonnegative, the first factor on the right in
(4. 1 5) is positive unless x = y = z = 0 . In order to demonstrate
(4. 14), it is sufficient to show that the second factor also is nonnega­
tive, i.e., that
(4. 16)
x2 + yz + z2 - xy - xz - yz :::::: 0 .
Referring back to the inequality
(x - y)2 = xz + yz - 2xy :::::: 0
already used in the algebraic proof of the arithmetic-mean-geometric­
mean inequality for two numbers [see subsection (b)], we see that the
inequality (4. 1 6) can be derived from this in the following fashion.
Write
(4. 1 7)
x2 + y2 :::::: 2xy ,
x2 + z2 :::::: 2xz ,
y2 + z2 :::::: 2yz ,
54
AN INTRODUCTION TO INEQUALITIES
and add the three inequalities. The result is
2(x2 + yz + z2 ) ;:::: 2(xy + xz + yz) ,
(4. 1 8)
which is equivalent to the desired inequality
equality holds if and only if x = y = z .
Since
(4. 16). The sign of
(4. 1 6) is a valid inequality, and since x + y + z ;:::: 0 , it fol­
lows that the left-hand side of (4. 1 5) is also ;:::: 0 ; that is, the inequality
(4. 14) is valid. Since (4. 14) is equivalent to (4. 1 1 ), we have now proved
that the arithmetic-mean-geometric-mean inequality is valid for
means of three numbers ; the condition x = y = z under which the
sign of equality holds in (4. 14), and therefore in (4. 1 1 ) , is equivalent
to the condition a = b = c .
(f)
The Arithmetic-mean-Geometric-mean Inequalityfor n Numbers.
Emboldened by this success, let us conj ecture that the results we
have obtained for means of two and three numbers are merely spe­
cial cases of a general theorem valid for any number of positive
quantities. If this conjecture is true, then we have the following result :
THEOREM 4.3. For any n nonnegative numbers a1 , a2 ,
•
.
.
,
a,.,
(4. 19)
The sign of equality holds if and only if a1
=
az
= · · · =
a,. .
This is the famous inequality connecting the arithmetic mean of n
quantities with the geometric mean of the n quantities, and is, indeed,
true. We have concentrated on this inequality for several reasons. In
the first place, it is a fascinating one, and one that can be established
in a large number of interesting ways; there are literally dozens of
different proofs based on ideas from a great variety of sources. In
the second place, it can be used as the fundamental theorem of the
theory of inequalities, the keystone on which many other very impor­
tant results rest. In the third place, as you will see in Chapter 5, we
can use some of its consequences to solve a number of maximization
and minimization problems.
In attempting to prove the general inequality, a first thought may
be to continue along the preceding lines, using another algebraic
factorization for n = 4 , still another one for n = 5 , and so on.
But this approach is neither attractive nor feasible. As a matter of
T H E CLASSICAL I N EQ U AL I T I ES
55
fact, no simple proof along these lines exists.
Instead, we shall present a simple proof based on two applications
of mathematical induction; one, a "forward" induction, will lead to
the desired result for all the integers
for
n
=
n
that are powers of two, i.e.,
2k ; the other, a "backward" induction (from any positive
in teger to the preceding one), together with the forward induction
will enable us to establish the result for all positive integers
(i)
n.
The method we shall now employ will
Forward Induction.
illustrate an amusing variant of the fundamental technique of proof
by mathematical induction, which we have previously discussed in
Sec. 2.6 (Chapter 2).
Let us start with the result for
a+
(4.20)
2
b
n
=
2 , namely
> y'ali
uu ,
_
·
which, by Theorem 4. 1 , is valid for all nonnegative a and b, and use
some mathematical ingenuity. Although there are very many simple
proofs of Theorem
ingenuity.
Set
where at.
values for
4.3, they all possess the common ingredient of
a2 , a3, a4 are nonnegative numbers. Substituting
a and b in (4.20), we obtain the inequality
a 1 + az a3 + a4
+
2
2
2
�
or
(4.2 1 )
a1 + az + a3 + a4
4
>
these
el i az)(a3 i a4) ·
J
j(al i az)e3 i a4) ·
Since the left-hand side of the inequality (4.2 1 ) has the desired form
(see Theorem 4.3), let us concentrate on the right-hand side. Using
the valid inequalities
(4.22)
56
AN I N T R O D U CT I ON TO IN E Q U A L I T I ES
2. 1), we obtain from (4.2 1) the
and the transitivity rule (Theorem
further inequality
a1 + az + a3 + a4
2::
(4.23)
4
� v�
a3a4
1 0
v v a 1a2
0
2:: (a 1a2a3a4)114 .
0
But this is precisely the desired result for four nonnegative quanti­
ties! The arithmetic mean is greater than or equal to the geometric
mean for n = 4 .
The sign of equality holds in
(4.2 1 ) if and only if
(4.22) if and only if a1 = a2 , a3 = a4 ; consequently, it holds
(4.23) if and only if a 1 = az = a3 = a4 .
and in
in
Nothing stops us from repeating the foregoing trick. Set
bi (i = I , 2, . . . , 8) are nonnegative quantities. Substitut­
(4.23), we have
where the
ing in
b 1 + bz
]
� • . . + bs 2:: [el i bz) b3 1 b�e5 1 b� 1 1 bH) 1 14
C
Using the inequalities
b1 + bz > n:;--r:y utuz ,
_
2
0
...
'
2
and the transitivity rule, we obtain
b1 + bz +
· · ·
8
-
b1 + bs >
�
V V7V8 ,
0
+ bs 2:: ( VJiJi; VliJii � Vfii�Js)1 14
2:: (btbz · · · bs)118 ,
the desired result for eight quantities. The sign of equality holds if
and only if all the
bi are equal.
Continuing in this way, we clearly can establish the inequality for
all values of n that are powers of two, i.e., for n = 2, 4, 8, 16, . . . .
To tie the result down rigorously, we use mathematical induction. The
main step consists of proving the following result:
The arithmetic-mean-geometric-mean inequality is valid for all n of
the form 2k, k = 1 , 2, . . . .
TH E CLASS ICAL I NEQUALITI ES
57
2 1, i.e.,
n=2=
2 2 and 23. Let us assume that the
result is true for an integer n of the form 2k, and then establish it for
2k+ 1 . Since 2k + 1 = 2 . 2k , this means that we shall prove that the
PRooF. We already know that the result is true for
for
k
=
I, and indeed for n
=
result is true for 2n.
Thus we have assumed that
(4.24)
for any set of nonnegative quantities at. a 2, . . . , a,. , where
Choose a; (i = I, 2, . . . , n) to have the following values:
n
=
2k .
where the 2n numbers b; U = I , 2, . . . , 2n) are given nonnegative
numbers, and substitute in (4.24). Proceeding as before, we finally
obtain
b1 + b2 + · · · + b2n
> (b1b2 · · · b2,.} 1 1211
2n
_
·
As before, the sign of equality holds if and only if all the
equal. Hence we have established the desired result for 2n, or
b;
are
2k+ 1 •
Thus, since the inequality is valid for k = I, the principle of
(forward) mathematical induction asserts that the result is true for
all positive integers k, and hence that the inequality
all n that are powers of two.
(ii)
Backward Induction.
(4.24) holds for
Now that we have established the
result for those integers that are powers of two, how do we establish
it for the full set of positive integers?
Another procedure is required. Consider the case
n
=
3 , for
which, of course, we have already established the result by a differ­
ent method. Using the relationship for n = 22 = 4 ,
(4.25)
a 1 + a2 + a3 + a4
4
> (a 1a2a3a4) 1 1 4 ,
_
which has already been established by forward induction, let us see
if we can derive the corresponding result for n = 3 .
58
AN I NT R0 D UCTI0N T0 IN EQ UA L ITIES
We accomplish this by means of the important technique of spe­
cialization. Starting with (4.25),
and a 4 in a special way. Set
we choose the quantities
a1,
a2, a3,
(4.26)
and ask for the value
a4 that yields the equality
a1 + a2 + a3 + a4
4
By the values given in
b1 + b2 + b3
3
(4.26), this requires that
whence
Substituting these particular values for the
relationship
ai
in
(4.25), we derive the
b1 + b2 + b3 > \¥1. lb1b2b3 ( b1 + b2 + b3)
3
3
.
Raising both sides to the fourth power, we obtain
(b1 + b2 + b3)4 >- b1b2b3 (b1 + b2 + b3)
(b1 + b2
,
(b1 + b2 + b3)3 >- b1b2b3 ·
3
+ b;>,)/ 3
or finally, dividing by
3
'
3
which is equivalent to the desired result
(4.27)
b1 + b2 + b3 >- V' b1b2b3 ·
Since equality holds in
3
b1 b2 b3•
(4.25) if and only if a1 = a2 = a3 = a4 , it
=
(4.27) if and only if
=
follows that equality holds in
In order to extend this method to the general case, we shall employ
an inductive technique, but an inductive technique of nonstandard
T H E C LASSICA L I N EQ UA LI TI ES
59
type. Instead of proving that if the result holds for n then it holds for
.
n
+ 1 , we shall prove that if it holds for n then it holds for n - 1 .
Since we already know that it holds for n = 2k (k = 1 , 2, . . ) , this
method will yield the theorem in all generality.
Let us now show that if the result holds for n, then it holds for
I . To do this, we repeat the trick of specialization that we used
above. Let
n -
(4.28)
0 •
and determine
•
'
an by the requirement that
a 1 + a2 + · · · + a,.
n
using the values
(4.29)
_
bt + b2 + · · · + b,._l .
'
n - I
(4.28) and solving for a,., we get
a,. =
bt + b2 + . . · + b,._t
.
n - I
We have assumed that the inequality
for the n nonnegative quantities
the values
a 1 , a2, . . . , an is valid; substituting
(4.28) and (4.29) for the a;, we have
_
bt + b2 +
n
" •
I
(
b 1 + b2 +
+ bn-1
n /b
> V
. 1b2 . . . bn- I
_
n
_
• "
I
+ bn-1
)
Raising both sides to the nth power and simplifying, we obtain the
inequality
( b! + b2 +
I
" •
n -
) _
+ bn- l n l
> b1b2 ' ' ' bn- 1 ,
which is equivalent to the desired result
b1 + b2 +
n
• • •
_
I
+ b,._ l
_
> "-::vlib1b2
As before, equality holds if and only if
the proof of Theorem
4.3 is complete.
• • •
bn-t .
b 1 = b2 = · · · = b,._ 1 , and
60
AN I N TROD UCTION TO I N EQ UALITI ES
4.3 Generalizations of the Arithmetic-mean­
Geometric-mean Inequality
We shall now show that a number of results that appear to be
generalizations of the fundamental arithmetic-mean-geometric-mean
theorem, derived above, are actually special cases.
First, let us take the arithmetic-mean-geometric-mean inequality
Xt
+
Xz
+
n
• • •
+ Xn :2:: XtXz
(
• • •
Xn)l ln
and set the first m of the numbers xi equal to the same nonnegative
value x, and the remaining n - m equal to a common nonnegative
value y; that is,
Xt = Xz =
• • •
Xm+l = Xm+2 =
= Xm = X ,
• • •
The arithmetic-mean-geometric-mean inequality for
becomes
or
n =y.
x 1 , xz, . . . , Xn
= X
mx + (n - m)y > x n m ln
- ( my - p
n
Here n is any positive integer, and m is any integer in the range
1 :::::; m :::::; n - I . It follows that min can represent any rational frac­
tion r occurring in the interval 0 < r < I Let us then write the
foregoing inequality in the form
.
(4.30)
rx + (I - r)y :2:: xry t -r,
a most important result for our subsequent purposes.
This inequality (4.30) is valid for any two nonnegative quantities
x and y, and for any fraction r between 0 and I . Equality occurs if
and only if x = y .
Let r be denoted by I !p . Since 0 < r < I , we see that p > I .
Then
I
-
r =
I
- .! = E._=l ,
p
p
Let q denote the quantity pl(p - 1 ) , so that 1 /q = I -
.!_ + .!_ = I .
p
q
r
and
T H E CLASSICAL IN EQU A LITIES
6I
The inequality (4.30) then has the form
(4.31)
� + l 2:: x llvyllq .
p
q
To eliminate the fractional powers, set
(4.32)
X = (JP ,
y
=
bq .
Then (4.3 1) assumes the form
(4.33)
aP
bq > ab
+-
q -
p
'
where a and b are nonnegative numbers, and p and q are rational num­
bers satisfying l ip + l !q = 1 . There is equality if and only if
(4.34)
Once we have defined what we mean by irrational numbers and
by functions of the form xr, where r is irrational, we can show,
either directly or by means of a limiting procedure starting from the
inequality (4.30), that the inequality (4.30) is actually valid for all r
between 0 and 1 , "'and therefore that (4.33) is valid for all p > 1 ,
q > I satisfying l ip + 1 /q = I . If you wish to pursue this refine­
ment further, you should first read Numbers: Rational and Irrational
by Ivan Niven, to which we have already referred in Chapter I .
Exercises
I. Show that if the values Yt.J2, . . . ,yk are all nonnegative, and if mt . m2 ,
. . . , mk are positive integers, then
ml)'t + mY'2 + . . . + m!Jik � (y m m2 • • • mk l l(m, + m. + • • • + m.>
yk )
t ':Y2
.
mt + m2 + · · · + mk
Show consequently that if r1, r2 , . . . , rk are proper fractions satisfying
r1 + r2 + · · · + rk = I ,
then
'l.Yt + r2y2 + · · · + rkyk � Yt''Y2'' Jk'•
·
· ·
·
2. An important average in statistics is the root-mean-square. For two non­
negative numbers a and b, the root-mean-square is the value
- j a2 + b2
s -
-2- .
For the pairs (5, 1 2), (0, 1 ), and (p. p), compute the arithmetic mean and
the root-mean-square.
62
A N I N T R O D U CTION TO I N EQ U AL I T I ES
3. Show that the arithmetic mean of two positive numbers is less than or equal
to their root-mean-square:
a+b<
a2 + b2
2 2
Under what circumstance does the sign of equality hold? How does the
root-mean-square compare with the geometric mean and with the har­
monic mean?
j
4. Let ABDC be a trapezoid with AB = a , CD = b (see Fig. 4.4). Let 0
be the point of intersection of its diagonals. Show that
Figure 4.4 Geometric Illustration of
2ab <
a+b
-
v!(ifuu <
-
.
a + b < V�
-y-2
-
(a) The anthmeuc mean (a + b) / 2 of a and b is represented by the line
segment GH parallel to the bases and halfway between them.
(b) The geometric mean yaTi is represented by the line segment KL
parallel to the bases and situated so that trapezoids ABLK and KLDC are
similar.
(c) The harmonic mean is represented by the line segment EF parallel to
the bases and passing through 0.
(d) The root-mean-square is represented by the line segment MN parallel
to the bases and dividing the trapezoid ABDC into two trapezoids of equal
area.
4.4 The Cauchy Inequality
(a) The Two-dimensional Version: (a2 + b2)(c2 + d 2) 2:: (ac + bd)2 .
Let us now introduce a new theme. As in a musical composition, this
theme will intertwine with the original theme to produce further and
more beautiful results.
We begin with the observation that the inequality
a2 + b2 2:: 2ab .
on which all the proofs in the preceding sections of this chapter were
based [see Sec. 4.2(b)], is a simple consequence of the identity
a2 - 2ab + b2
=
(a - b)2 ,
T H E CLASSICAL I N EQ U A LIT I ES
63
which is valid for all real a, b, not merely for nonnegative a, b.
Consider now the product
(4.35)
(a2 + b2)(c2 + d 2 ) .
We see, upon multiplying out, that this product yields
a2c2 + b2d 2 + a2d 2 + b2c2 ,
which is identically what we obtain from expanding the expression
(4.36)
(ac + bd)2 + (be - ad)2 .
Hence we have
(a2 + b2 ) (c2 + d 2 ) = (ac + bd ) 2 + (be - ad )2 .
Since the square (be - ad )2 is nonnegative, from (4.37) we obtain
(4.38)
(a2 + b2) (c2 + d 2) 2:: (ac + bd )2 ,
for all real a, b, c, d,
(4.37)
a very pretty inequality of great importance throughout much of
analysis and mathematical physics. It is called the Cauchy inequality,
or, more precisely, the two-dimensional version of the Cauchy
inequality.t
Furthermore, we see from (4.37) that the sign of equality holds in
(4.38) if and only if
(4.39)
be - ad = 0 .
In this case, we say that the two pairs (a, b) and (c, d ) are proportional
to each other; if c =¥= 0 and d =¥= 0 , the condition (4.39) can be
written as
(b) Geometric Interpretation. On first seeing the identity of the
expressions (4.35) and (4.36), the reader should quite naturally and
legitimately wonder how in the world anyone would ever stumble
upon this result. It strikes him as having been "pulled out of a hat,"
a piece of mathematical sleight of hand.
It is a tenet of a mathematician's faith that there are no accidents
in mathematics. Every result of any significance has an explanation
t A generahzation of this inequality to expressions occurnng in integral calculus
was discovered independently by the mathematicians Buniakowski and Schwarz. The
name "Cauchy-Schwarz inequality" is sometimes applied to the inequality in the text,
but more particularly to its generalization.
64
AN I N TRODUCTION TO IN EQ UALIT ES
which, once grasped, makes the result self-evident. The explanation
may not immediately be obvious, and it may not be found for some
time. Often the significance of a mathematical theorem becomes clear
only when looked at from above-that is to say, from the standpoint
of a more advanced theory. But the meaning is always there. This is
a vitally important point. Were it not for this, mathematics would
degenerate into a collection of unrelated formalisms and parlor tricks.
Frequently, the simplest interpretation of an algebraic result is in
terms of a geometric setting. Formulas that seem quite strange and
complex become obvious when their geometric origin is laid bare.
y
Figure 4.5 Geometnc interpretation of the Cauchy inequality
Consider the triangle shown in Fig. 4.5. The lengths of the seg­
mentst OP, OQ, and PQ are given by
(a2 + b2)1 12 '
OQ = (c2 + d 2) 1 1 2 ,
OP
=
and
PQ
=
[(a - c)2
+
(b - d)2] 1 12 ,
respectively.
Denote the angle between OP and OQ by 0. By the law of cosines,
we have
(PQ)2
=
(OP)2 + (OQ)2 - 2( 0P)(OQ) cos 0 .
Substituting the values of OP, OQ, and PQ, and simplifying, we
obtain
(4.40)
cos 0
_
-
ac + bd
+
2
2
b )1 12 (c2 + d 2) 1 12
(a
t The length of a segment X Y, often written X Y, will be wntten X Y m this book for
reasons of typography
65
T H E C LASSICAL I N EQ U A L IT I ES
Since the cosine of an angle must always lie between - 1 and + 1 ,
we must have
0 :::::; cos20 :::::; 1 .
(4.4 1)
Squaring both sides of (4.40) and (4.4 1 ), we obtain
cos20 _
and finally
(ac + bd )2
:::::; 1 ,
2
+
b2)(c2 + d 2 )
(a
(a2 + b2 )( c2
+
d 2) 2:: (ac + bd)2 .
This is again the two-dimensional version of Cauchy's inequality
(4.38), which seemed so magical in its algebraic setting.
Furthermore, we see that there is equality if and only if cos20 = 1 ,
i.e., if and only if 0 is a zero or straight angle-that is to say, if and
only if the points 0, P, and Q lie on the same line. In that case, we
must have an equality of slopes; in other words, unless c = d = 0 ,
(c) Three-Dimensional Version of the Cauchy Inequality. An advan­
tage of an interpretation of the foregoing type lies in the fact that we
can use geometric intuition to obtain similar results in any number
of dimensions.
Turning to three dimensions, let P : (at . a2 , a3) and Q : (b1 , b2 , b3)
be points distinct from the origin 0 : (0, 0, 0) . Then the cosine of the
angle 0 between OP and OQ is givent by
cos 0
=
(a1 2 + a22
a1b1 + a2b2 + a3b3
+ a3 2) 1 12 (b 1 2 + b 2
2
+
b32) 1 12 '
which, together with the fact that cos20 :::::; 1 , implies the three­
dimensional version of the famous inequality of Cauchy :
(4.42)
(a1 2 + a22 + a32)(b1 2 + b2 2 + b32) 2:: (a1b1 + a2b2
+
a3b3)2 .
The sign of equality holds if and only if the three points 0, P, and
t For a derivation see W. F. Osgood and W. C. Graustein, Plane and Solid A narytic
Geometry, Macmillan and Co., New York, 1 930.
66
AN INTRODUCTION TO IN EQUALITIES
Q are collinear; this is expressed by
a1
bl
a3
,
b3
a2
b2
provided the b 's are all different from zero.
(d) The Cauchy-Lagrange Identity and the n-Dimensional Version of
the Cauchy Inequality. A strictly algebraic proof of the three-dimen­
sional Cauchy inequality (4.42) may be obtained by noting that
(a1 2 + a2 2 + a3 2 )(h1 2 + b22 + b3 2 ) - (a1h1 + a2b2 + a3b3)2
= (a1 2b 22 + a22b 1 2) + (a 1 2b32 + a3 2b 1 2 ) + (a 22h3 2 + a32b22)
( 4.43)
- 2a1b1a2b2 - 2a1b1a3b3 - 2a2b2a3b3
= (a1b2 - a2b1)2 + (a1b3 - a3b1) 2 + (a2b3 - a3b2 )2 .
The last expression in (4.43) clearly must be nonnegative, since it
is a sum of three nonnegative terms. Hence
(a1 2 + a22
+
a32 )(b1 2 + b22
+
b32 ) - (a1b1 + a2b2 + a3b3)2 � 0 ,
and thus Cauchy's inequality for three dimensions is proved again.
The identity (4.43) yields both the inequality and the case of equality;
the last expression in (4.43) is zero if each term vanishes, i.e., if the
a 's and b's are proportional .
When you study the analytic geometry of three dimensions, you
will see that the identity (4.43) is simply the result
cos20 + sin 20
=
1
written in another way.
The identity (4.43) can be generalized as follows: For any set of
real quantities a; and b; (i = 1 , 2,
, n) , we have
.
.
.
··· +
an2)(b1 2 + b22 + · · · + bn2)
(4.44)
- (a1b1 + a2b2 + • • • + anbn)2
= (a1b2 - a2b1)2 + (a1b3 - a3b1) 2 + • • • + (an-l bn - anbn-1)2 ;
this is the famous identity of Cauchy and Lagrange. From this identity
(a1 2 + a22
+
we obtain the n-dimensional version of the Cauchy inequality
(4.45)
(a1 2 + a22 + · · · + an2)(b1 2 + b22 + · · · + bn2)
� (a1b1 + a2b2 +
valid for all real values of the a; and b;.
•••
+
anbn)2 ,
T H E C LASS ICAL IN EQUALITI ES
67
(e) A n A lternative Proof of the Three-Dimensional Version. The
proofs that we have given above were perfectly satisfactory as far as
the results at hand were concerned. However, they do not generalize
in such a way as to yield some results that we wish to derive later on.
Consequently, we start all over again and introduce a new device.
Let us begin with our basic inequality (x - y ) 2 2:: 0 in the form
x2 y2
xy
2 + 2 2:: ,
(4.46)
which is valid for all real x and y, and substitute, in turn,
x=
then
a1
,
(a 1 2 + a22 + a32 )1 12
y=
b1
,
(b1 2 + b22 + b32 ) 1 12
then
where the a; and b; are all real quantities. Adding the three inequali­
ties obtained in this way, we find
) (
)
l b1 2 + b2 2 + b32
l (a12 + a22 + a32
"2\a 12 a22 a:12 + 2 b12 b22 b32
+
+
+
+
a1b 1 + a2b2 + a3b3
>
,
- (a 1 2 + a22 + a32) 1 12(b1 2 + b22 + b3 2 )1 1 2
which, very obligingly, yields the inequality
1 >
- (a1 2
+
a2 2
a1b1 + a2b2 + a3b3
+ a3 2) 1 12(b 2
1 + b2 2
+
b3 2) 1 12 '
equivalent to Cauchy's inequality (4.42) [subsection (c)], as desired.
The n-dimensional version (4.45) of Cauchy's inequality may be
proved in an analogous manner.
4.5 The Holder Inequality
We now possess all the tools necessary to fashion one of the most
useful inequalities of analysis, the HOlder inequality. This states that
68
AN I N T R O DU CTION T O I N EQ U A LITIES
for any set of nonnegative quantities
have
(4.47)
(alP + azP +
···
ai and bi (i =
I, 2,. . . , n) , we
+ anP) 1 1P(bl q + b2q + · • · + hnq)l lq
2:: a1b1 + a2b2 + · · · + anbn ,
where p and q are related by the equation
(4.48)
and p > 1 .
The case p = q = 2 is the Cauchy inequality established in the
preceding sections. In the general case, however, we must restrict our
attention to nonnegative ai and bi, since fractional powers p and q
might be involved.
Actually we shall prove (4.47) only for rational numbers p, q; how­
ever, the result is valid also for irrational p and q.
We begin with the inequality
aP + bq > ab
p
q - '
which we established in Sec. 4.3 ; see (4.33) for rational p, q and non­
negative a, b. Then we use the trick employed in Sec. 4.4. We set
a1
a_
+
P
azP
(a
+ . . . + anP) l lp '
- 1
then
and so on, and add the resulting inequalities. We thus obtain
I f.a1P
(4.49)
+ azP +
p\alP + a2P +
•••
•··
)
+ anP + I(b 1q + b2q + · · • + bnq
+ anP q\blq + b2q + · · · + bnq
)
alb ! + a2b2 + . . . + anbn
>
- (a lP + azP + • • • + anP)1 1P(blq + b2q + • • • + bnq)l lq '
or, finally, using (4.48), the result stated in the inequality (4.47).
Equality can occur in ( 4.49) if and only if biq/aiP has a common
value for all i.
T H E C LASSICAL I N EQ U A LITIES
69
4.6 The Triangle Inequality
We are all familiar with the geometric theorem that the sum of the
lengths of two sides of a triangle is greater than or equal to the length
of the third side. Let us see what this implies in algebraic terms.
Take the triangle to be situated as shown in Fig. 4.6. Then the
geometric inequality
OP + PR
:2:
OR
is equivalent to the algebraic triangle inequality
(4.50)
yx1 2 + y12 + yx22 + y22
:2:
y(x 1 + x2) 2 + (yl + y2)2 .
y
Figure 4 6 The tnangle inequality
Can we establish the triangle inequality (4.50) without resorting to
geometry? In the one-dimensional case this was done in Sec. 3.8 (see
Theorem 3.2), where the inequality was written in the form
I x 1 l + I x2 l :2: I X1
+
x2 l
more often than in the equivalent form
N + N :2:
y(x1 + x2)2 .
Now, the simplest way of proving the two-dimensional triangle
inequality (4.50) is to establish the validity of an equivalent inequality.
To this end, we square both members of (4.50), obtaining the equiva­
lent inequality
x1 2 + Y12 + X22 + Y22 + 2yx1 2 + Y1 2 yx22 + Y22
:2: (x1 + x2)2 + (y1 + y2)2 ,
which is readily seen to be equivalent to
(4.5 1)
yx1 2 + Y1 2 yx22 + y22 :2: X 1X2 + Y1Y2 ;
70
AN I NT R0 D UCTI 0N T0 I N EQU ALITI ES
but this is a simple consequence of the familiar Cauchy inequality
[two-dimensional version, see (4.38)]
(4.52)
and thus the triangle inequality is established.
As in the one-dimensional case (Theorem 3.2), the discussion of
the conditions under which the sign of equality holds in the triangle
inequality (4.50) is just a bit involved. You will recall that in the
Cauchy inequality (4.52) the sign of equality holds if and only if
(x1.y1 ) and (x2 ,y2) are proportional, i.e., x1 = kx2 , Y 1 = ky2 . Now
(4.5 1 ) is obtained from (4.52) by taking a square root of each mem­
ber; this is a valid operation since it is the nonnegative square root of
the left-hand member that was taken. But if x1x2 + Y:tY2 < 0 ,
that is to say, if x1x2 + yl)'2 is the negative square root of the
right-hand member (x1x2 + yl)'2 ) 2 of (4.52), then the strict inequality
holds in (4.5 1 ) even though (x1.y1) and (x2 ,y2) are proportional. Thus
the sign of equality holds in (4.5 1 ), and hence also in the triangle in­
equality (4.50), if and only if X1 = kx2 and Yl
negative constant k ofproportionality. t
=
ky2 with a non­
Geometrically, the foregoing necessary and sufficient condition for
the sign of equality to hold in the triangle inequality (4.50) has the
interpretation that the points 0, P, and Q in Fig. 4.5 be collinear and
that P and Q be on the same side of 0. Then the triangle OPR col­
lapses. In this case, we say not merely that P and Q are collinear with
0, but that they are on the same ray issuing from 0.
You might check that the foregoing discussion is consistent with
the discussion of the corresponding condition in the one-dimensional
case I a I + I b I 2 I a + b I , namely, that the equality sign holds if and
only if a and b are of the same sign.
Our proof of the triangle inequality can be generalized, following
the pattern set in the proof of the Holder inequality, to yield
\fx1 2 + X22 + • • • + Xn2 + VY 1 2 + Y2 2 + • • • + Yn2
2 y(x l + Y 1)2 + (x2 + Y2 )2 + • • • + (Xn + Yn)2 ,
valid for all real xi, Yi. and with the sign of equality holding, as before,
if and only if the Xi and the Yi are proportional with a positive factor
of proportionality. We shall return to this inequality in Chapter 6,
t For example, ( - 3, 4) and (6, - 8) are proportional with a negative constant of
proportionality, while ( -3, 4) and ( - 6, 8) have a positive constant of proportionality.
71
T H E C LASSICAL I N EQ U A LITIES
when we consider its geometric significance.
An alternative proof of the triangle inequality, which can be
extended to give a more general result, is the following. We write the
identity
(x1
x2)2 + (yl + Y2)2
= X l (X l + X 2) + Y (Y l
l
+
+
Y2)
+
X2(Xl
+
X2)
+
Y2(Yl
+
Y2)
and apply the Cauchy inequality in square-root form [see eq. (4.5 1)]
separately to the two expressions
x 1(x 1
+
x2)
+
y1 (Y1
+
y2 )
and
obtaining
+
(x1 2
Y1 2)1 1 2[(xl
+
X2)2
(yl
+
Y2)2 )1 12
:;::: X!(X l
+
X 2)
Y l(Y l
+
Y2) '
(x2 2 + Y2 2)1 1 2((xl
+
x2)2 + (yl
+
Y2)2) 1 12
:;::: X2(Xl
+
X2) + Y2(Yl
+
Y2) .
+
+
and
Adding, we have
+
y 1 2) 1 12
+
+
y2 2) 1 12 ) [(x 1
+
x2)2 + (y1 + y2)2)1 12
:;::: (x 1 + x2)2 + (yl + y2)2 .
Dividing through by the common factor [(x 1 + x2)2 + (y1 + y2)2] 1 12 ,
[(x 1 2
(x22
we get
(x1 2 + y12) 112
+
(x22 + y22) 1 1 2 :;::: [(x1
+
x2)2
+
(y 1
+
y2)2) 1 12 .
Thus we have again established the triangle inequality (4.50).
Returning to the point where the Cauchy inequality was applied
in square-root form, we see again that equality occurs if and only if
for some nonnegative factor of proportionality k-that is to say, if
and only if the three points 0, P, and Q are collinear, with P and Q
on the same side of 0.
72
A N I NT R O DU CTION TO I N EQ U A LITIES
4.7 The Minkowski Inequality
We now possess all the tools and devices to establish another
famous inequality, that due to Minkowski. It asserts that for any set
of nonnegativet quantities X t , Yt • x 2 , y2 , and any p > I , we have
(4.53)
(x 1P + y 1P)1 1P + (x� + y�)liP
�
[(Xt + X2)P + (Yt + Y2)P) 11P .
The triangle inequality is the special case p = 2 .
The proof of Minkowski's inequality is similar to that just given
for the triangle inequality, with the difference that the more general
Holder inequality (Sec. 4.5) is used in place of Cauchy's inequality.
Write the identity
(Xt + X2)P + (yl + Y2)P = [X t(X t + X2)P- l + Yt(yt + Y2)P- l )
+ [x2(x 1 + x2)P- 1 + Y2(Yt + Y2)P- 1 ]
and apply Holder's inequality to each of the terms separately. The
results are
(XtP + YtP)l iP((Xt + X2)(p- l)q + (Jt + Y2)(p-llq) llq
� Xt(Xt + X2)p- l + Yt(Yt + Y2)P- l
and
(x� + y�)l l�>[(xt + x2)(11- tlq + (Yt
�
Since 1 /p + I /q
=
+ Y2)(11- tlq ) llq
X2(Xt + X2)p- l + J2(yl
I , we see that (p - I )q
=
+ Y2)P-l .
p . Adding, we obtain
[(x t + x2)P + (Yt + y2)P]1 1Q[(x 1P + y1P) 1 1P + (x� + y�)l iP)
� (Xt + X2)P + (yl + Y2)P ;
and then dividing by [(xt + x2)P + (y 1 + y2)P] 11q , we get
(x tP + YtP)1 1P + (x2P + y2P)1 1P � [(x t + x2)P + (Yt + y2)P)1 - 1 1q
which, since
(4.53).
I - l lq
=
I !p , is the stated Minkowski inequality
The sign of equality holds in the Minkowski inequality if and only
if it holds in the Holder inequality [by means of which (4.53) was
proved], that is, if and only if the points ( x 1, y 1 ) and (x2 ,y2) [which
are in the first quadrant] are collinear.
t The restriction to nonnegative quantities is necessary because, in general, fractional
powers p and q are involved.
73
T H E C LASSICAL I N E Q U A LI T I ES
As you might surmise from the foregoing generalizations of the
Cauchy, Holder, and triangle inequalities, the general Minkowski
inequality for sets of n nonnegative numbers
Xt. X2,
•
.
.
, Xn
and
J t, J2, . •Yn
.
·
is
[x 1P + x� + . . . + XnP) l iP + [y 1P + y2P + . . . + YnPJ l iP
:2: [(Xt + Yt )P + (x2 + y2)P + • • • + (Xn + Yn)P)1 1P
for p ;;::: 1 . The inequality sign is reversed for p < 1 .
4.8 Absolute Values and the Classical Inequalities
The arithmetic-mean-geometric-mean inequality, the Cauchy in­
equality, the Holder inequality, the triangle inequality, and the
Minkowski inequality are the classical inequalities of mathematical
analysis. For ready reference, they are collected in Table 3.
TABLE 3. The Classical Inequalities for Nonnegative Values
Inequality
Name
Arithmetic-mean­
geometric-mean
inequality
Cauchy
inequality
(a1 2 + a2 2 + . . . + a,.2)112(b1 2 + b22 + . . . + b,.2)1 12
� a1b1 + a2b2 +
+ a,.b,.
Holder
inequality
(a1P + a2P +
Triangle
inequality
(a12 + a22 + . . . + a,.2) 112 + (b1 2 + b22 + . . . + b,.2)112
� ((a1 + b1)2 + (a2 + b2)2 +
+ (a,. + b,.)2)1 12
Minkowski
inequality
(a1P + a2" + . . . + a,.P)liP + (b1P + b2" + . . . + b,.P)liP
� ((a1 + b1)" + (a2 + b2)" + · · · + (a,. + b,.)"]1 1P
· · ·
• • •
+ a,.P)11P(b1q + b2q +
� a1b1 + a2b2 +
• • •
+ b,.q)llq
+ a,.b,.
· · ·
· · ·
, an
The inequalities are valid for any nonnegative values a1 , a2,
and bt, b2 , . , bn; for arbitrary p > 1 ; and for 1 /p + 1 /q = 1 .
The Cauchy inequality and the triangle inequality constitute the
special case p = 2 of the Holder inequality and the Minkowski
•
.
.
•
•
AN I NTR0 D U CTI 0N T0 I N EQU A L IT I ES
74
inequality, respectively. The sign of equality holds if and only if all
the a; are equal, for the arithmetic-mean-geometric-mean inequality;
the sets (a;�>), (b;q) are proportional, for the Holder inequality; and the
sets (a;), (b;) are proportional, for the other inequalities.
The foregoing inequalities are primarily concerned with nonnega­
tive values. However, the absolute value of any real number is non­
negative, so that the inequalities apply, in particular, to the absolute
values of arbitrary real numbers. This observation can be extended
somewhat by means of the result in Theorem 3.2, which states that
a sum of absolute values is greater than or equal to the absolute value
of the corresponding sum; thus, relative to the Minkowski inequality,
we have
the sign of equality holding if and only if a; and b; are of the same
sign. The extended inequalities are exhibited in Table 4.
TABLE 4. The Classical Inequalities for Arbitrary Values
Name
Inequality
Arithmetic-mean­
geometric-mean
inequality
Cauchy
inequality
Holder
inequality
Triangle
inequality
Minkowski
inequality
(a 1 2 + a22 +
. • •
+ a,.2)1 12(b1 2 + b22 + + b,.2) 1 12
2 a1 b1 + a2b2 + · · · + a,.b,.
. • .
+ I a2 i " + • • • + I a,. I ") 1 1"
< I bt l q + I b2 l q + • • • + I b.. I qptq
2 a1 b 1 + a2b2 + · · · + a,.b,.
< I a1 l "
•
(a1 2 + a22 + . . . + a,.2) 1 12 + (b1 2 + b22 + . . . + b.. 2) 11 2
2 [(at + bt)2 + (a2 + b2)2 + · + (a,. + b,.)2J1 12
• •
( I at I " +
2
l a2 I P + · · · + l an i " ) 1 1"
+ < I bt l " + I b2 i " + · · · + I b,. I "P 1"
[ ! at + bt l " + l a2 + b2 i " + · · · + i an + b,. I PJl'P
T H E CLASSICAL INEQUALITIES
75
For these inequalities involving arbitrary real values a1 , a2, • . . , an
and bt . b2 , . . . , bn, the sign of equality holds if and only if all the 1 ai I
are equal, for the arithmetic-mean-geometric-mean inequality; the
sets ( I ai 1 �>), ( I bi I q) are proportional and each ai bi nonnegative, for the
Holder inequality; and the sets (ai), (bi) are nonnegatively propor­
tional. for the other inequalities.
As an application, for given sets of values (x 1 , x2 , • • • , Xn),
(y1 .y2, . . . , yn), and (zt , z2 , . . . , Zn), let
Then
ai = zi - Yi
and
and the extended Minkowski inequality yields
+ l zn - Yn l 11)1111
( l zt - Yt l 11 + 1 z2 - Y2 l 11 +
+ ( IYt - Xt l 11 + IY2 - X2 l 11 +
+ IYn - Xn l 11) 1 111
� ( l zt - Xt l 11 + l z2 - X 2 l 11 +
+ l zn - Xn l 11)1111 ;
• • •
• • •
• • •
equality holds if and only if the sets (zt - Yt . z2 - y2 , . . . , Zn - Yn)
and ( Yt - X t.. Y2 - x2, . . . ,yn - Xn) are proportional with a non­
negative constant of proportionality.
4.9 Symmetric Means
You might think, by this time, that the subject of inequalities has
been pretty well exhausted. Actually, j ust the opposite is true. All
that has gone before represents the barest scratching of the surface.
The inequalities that we have established can be generalized in count­
less ways, and then these generalizations in turn can be generalized.
They can be combined in many ingenious ways to give many further
results, as we shall indicate in this section and in Sec. 4. 10.
For example, one extension of the arithmetic-mean-geometric-mean
inequality that is rather interesting is the following: Form the three
means,
mt
=
Xt + X2 + X3 ,
3
( XtX2 + XtX3 + X2X3 �l/2
m2 _
3
rna
=
(XtX2Xa) 1 13 •
)
,
76
A N I N TR0 D U CTI 0N T0 I N EQ U A LITI ES
Then the continued inequality
mt � m2 � m3
is valid for all nonnegative quantities X t . x 2 , and x 3• We leave the
proof of this for the reader. This, in turn, is a special case of the analo­
gous result valid for any n nonnegative quantities, in which there are n
mean values starting with the arithmetic mean and ending with the
geometric.
4.10 The Arithmetic-Geometric Mean of Gauss
Let a and b be two nonnegative quantities, and consider the
quantities a1 and b1 defined in terms of a and b by
By our basic Axiom II, page 7, a 1 and b1 must also be nonnegative.
If we assume that a > b , we have
a + a a,
at < 2- =
Furthermore, by the arithmetic-mean-geometric-mean inequality,
a 1 > bt .
Let us now repeat this process, introducing a 2 and b2 by means of
the formulas
Then, by the same reasoning
as
above,
a > a1 > a2 ,
b < bt < b2 ,
and
a2 > b2 .
Let us now continue this process, defining
77
T H E C LASSICAL I N EQ U A L I T I E S
then a4, b4, and generally an. bn. by the recurrence relationships
On-1 + bn - 1
,
On =
2
(4.54)
After the kth step of our process, we have numbers at , a2 ,
b 1 , b2 , . . . , bk satisfying the inequalities
•
•
•
, ak and
For example, if a = 4 and b = 1 , then the first few a's and b's are
represented on the line shown in Fig. 4.7. We observe that b is the
smallest and a the largest of all these numbers, and that all the b's
are smaller than all the a's.
2
0
3
4
Figure 4.7. The anthmetic-geometric mean of Gauss
We need not stop after k steps. Suppose we continue defining more
and more a's and b's by the relationships (4.54). Each pair at. b, so
defined, is sandwiched between the previous pair ai-t. bi_1• Thus it
is plausible that the quantities an. which decrease as n gets larger but
which must always remain bigger than each bi , approach some
fixed value A ; similarly, the bn, which increase as n gets larger but
which remain smaller than each at. approach a fixed value B. The
reader familiar with the concept of limit will readily see that A is the
limit of the infinite sequence of numbers {an } and B is the limit of
the infinite sequence { bn } ·
Moreover, the difference an - bn becomes rapidly smaller a s n
increases. We can show that, at each step, this difference is less than
half of what it was at the previous step. Just write, using (4.54),
(4.55)
Since
+ bn
an+1 - bn+1 - On
2
Vfi:< yo:,
r;;,:;- _
V Onvn -
_
_
-
we have
2bn < 2 ya;:Fi;. ;
an - 2 -VaJi': + bn
2
·
78
adding
AN I NTR0DU CTI0N T0 IN EQU ALITI ES
an
-
bn
-
2 � to both sides of this inequality, we obtain
an
-
2 .....;a;Ji;. + bn < an - bn
which, when applied to (4.55), yields
as we set out to show.
Since the a's and b's approach each other in this manner, their
limits A and B must coincide. So
A = B,
and this common limit depends only on the numbers a and b with
which we started; in mathematical language, A is a function of a and
b. It was shown by Gauss that this function is not just a curiosity but is
rather basic in analysis ; it may be used to found the branch of
mathematics called the "theory of elliptic functions."
CHAPTER FIVE
Maximization and Minimization
Problems
5.1 Introduction
Now we shall demonstrate how the inequalities that were derived
in the preceding chapter can be used to treat an important and fasci­
nating set of problems. These are maximization and minimization
problems.
In the study of algebra and trigonometry, the problems you encoun­
tered were all of the following general nature : Given certain initial
data, and certain operations that were to be performed, you were to
determine the outcome. Or, conversely, given the operations that were
performed and the outcome, you were to determine what the initial
data must have been.
For example, you met three workmen, eager and industrious A,
plodding and conscientious B, and downright lazy C. They were put
to work digging ditches, constructing swimming pools, or building
houses, and the problem was always that of determining how long it
would take the three of them together to perform the job, given their
individual efficiency ratings-or, given the time required for all three
together, and the rates of effort of two of the three workers, it was
required to find the rate of the third.
Sometimes you were given triangles in which two sides and an
included angle were known, and sometimes you were given three sides
79
80
A N I N T RO D U CTION TO I N EQUALITIES
of the triangle. In each case, the problem was that of determining the
remaining sides and angles.
The foregoing kinds of problems are simple versions of what may
be called descriptive problems.
Here, we wish to consider quite different types of problems. We
shall consider situations in which there are many ways of proceeding,
and for which the problem is that of determining optimal choices.
Questions of this nature occur in all branches of science and consti­
tute one of the most important applications of mathematical analysis.
Furthermore, much of physics and engineering is dominated by prin­
ciples asserting that physical processes occur in nature as if some
quantity, such as time or energy, were being minimized or maximized.
M any processes of this sort can be treated in routine fashion by
that magic technique of Fermat, Leibniz, and Newton, the calculus.
Supposedly, in the 1 7th century, men who knew calculus were pointed
out in the street as possessing this extraordinary knowledge. Today
it is a subject that can be taught in high school and college, and
requires no exceptional aptitude.
When you take a course in calculus, you will find that many of the
problems treated here by algebraic processes can also be solved quite
readily by means of calculus. I t will then be amusing to solve them
mentally by means of the techniques presented here, but to follow
the formalism of calculus in order to check your answer.
Each technique, calculus as well as the theory of inequalities, has
its own advantages and disadvantages as far as applications to maxi­
mization and minimization problems are concerned. It is rather
characteristic of mathematics that there should be a great overlapping
of techniques in the solution of any particular problem. Usually, a
problem that can be solved in one way can also be solved in a num­
ber of other ways.
5.2 The Problem of Dido
According to legend, the city of Carthage was founded by Dido, a
princess from the land of Tyre. Seeking land for this new settlement,
she obtained a grudging concession from the local natives to occupy
as much land as could be encompassed by a cowhide. Realizing that,
taken literally, this would result in a certain amount of overcrowding,
she very ingeniously cut the skin into thin strips and then strung them
out so as to surround a much larger territory than the one-cowhide
area that the natives had intended.
M A X I M I Z AT I O N A N D M I N I M I Z AT I O N
81
The mathematical problem that she faced was that of determining
the closed curve of fixed perimeter that would surround the greatest
area; see Fig. 5 . 1 .
I n its general form, the problem is much too complicated for u s to
handle. As a matter of fact, at our present level we don't even know
how to formulate it in precise terms, since we have no way of express­
ing what we mean either by the perimeter of an arbitrary curve or
by the area enclosed by this curve. Calculus suggests definitions and
provides analytic expressions for these quantities, and a still more
advanced part of mathematics called the calculus of variations pro­
vides a solution to the problem.
As one might guess, the optimal curve in Dido's problem is a
circle: Let us here, however, concentrate on certain simple versions
of the problem that we can' easily handle by means of the fundamen­
tal results we have already obtained in the theory of inequalities. If
you would like to pursue the study still further, you should read the
interesting tract Geometric Inequalities by N . D. Kazarinoff, also in
this series.
5.3 A Simplified Version of Dido's Problem
Let us suppose, for various practical considerations, that Dido was
constrained to select a rectangular plot of land, as indicated in Fig. 5.2.
X
X
Figure 5 . 1 .
Figure 5.2.
The problem o f Dido
Simplified version of the problem of Dido
With the lengths of the sides of the rectangle designated by x and
y, respectively, we see that the length of its perimeter is given by the
algebraic expression
(5. 1)
L = 2x + 2y ,
and its area is represented by the formula
(5.2)
A = xy .
82
AN I NT R0D U CTI 0N T0 IN EQ U AL lTI ES
Since x andy represent lengths, they necessarily are nonnegative
quantities. The fact that the perimeter 2x + y
2 is equal to L implies
that x andy must satisfy the inequalities
(5.3)
L
2
� X � 0'
L
T
�y � 0 .
It is clear, then, that the area A = xy cannot be arbitrarily large.
Indeed, from the inequalities (5.3) and the expression (5.2) for the
area, we see by Theorem 2. 5 of Chapter 2 that A cannot exceed the
value L2 14 ; that is, we have
(5.4)
L2
4 � xy= A .
How are we going to determine the dimensions that yield the
maximum area?
Referring to the inequality connecting the arithmetic mean and
geometric mean of two quantities, we observe that
(5.5)
for all nonnegative numbers x andy. Since, in this case, we have
x +y= L/ 2 , the inequality (5.5) yields the relationship
(5.6)
or
L2
� xy= A .
16
Consequently, we see that our first rough bound for the area, L214 ,
obtained in (5.4), can be considerably "tightened." We can, however,
go much further. Recall that we have established that the equality in
(5.6) holds if and only if x = y . In our case, this means that the
new upper bound, L2I 16 , is attained if and only if x= y . For all
other nonnegative choices of x andy satisfying the relationship (5. 1 ),
the area is less than L2I 1 6 .
What does the foregoing analysis tell us? It tells us that in no case
can the area of a rectangle with perimeter L exceed the quantity
L2I 1 6 , and that this maximum value is actually reached if and only
if we choose the sides equal to each other, and therefore equal to the
quantity L14 .
Thus, by purely algebraic means we have achieved a proof of the
well-known and rather intuitively obvious fact that the rectangle of
greatest area for a given perimeter is a square.
83
M A X I M I ZATION A N D M I N I MIZATI ON
5.4 The Reverse Problem
Let us now consider the problem of determining the rectangle of
minimum perimeter that encloses a fixed area. This is a
reverse problem to the original.
Returning to the equations
dual
or
(5. 1 ) and (5.2), we see that we now wish
to determine the nonnegative values of x andy that make the expres­
sion 2x + 2y a minimum, while preserving a fixed value
product xy.
A
for the
As might be expected, the same inequality between the arithmetic
and geometric means yields the solution to this problem. From the
relationship
(5.5) above, we know that
(x + y)2
4
>
xv
= A .
'/
-
Hence we see that
and that
L
=
2x + 2y � 4 y'A.
Consequently, we can assert that the perimeter must be at least as
great as
4 VA
and, furthermore, that this minimum value is actually
attained if and only if x = y =
y'A.
Hence, once again, the optimal
rectangular shape must be that of a square.
This reciprocal relationship between the solutions of the two prob­
lems is no accident. Usually, in the study of variational problems of
this type, the solution of one problem automatically yields the solu­
tion of the dual problem as well. For a proof of this duality principle,
see, e.g., N. D. Kazarinoff's
Geometric Inequalities mentioned earlier.
5.5 The Path of a Ray of Light
Suppose we wish to determine the path of a ray of light going from
a point P to a point Q by way of reflection in a plane surface, as shown
in Fig.
5.3. Actually, the problem as here stated is a three-dimensional
one, but an extension of the following analysis shows that the ray
must travel in the plane through P and Q and perpendicular to the
reflecting plane.
We assume that the medium is homogeneous, so that a ray of light
travels at constant speed. How shall we determine the point
R
and
84
AN INTRODUCTION TO INEQUALITIES
the paths PR and RQ? Let us invoke Fermat's principle, which states
that the total length of time required must be a minimum for all pos­
sible choices of the point R. Again, because the medium is homoge­
neous, this means that the paths PR and RQ are straight lines and
that R is so located that the length PR + RQ is a minimum.
Let the coordinates of P, Q be (0, a), (q, b) respectively, and let r
be the unknown distance OR. Then
OP = a,
and
RT = q - r,
OR = r,
TQ = b,
PR = y'a2 + r2 ,
RQ =
y'b2 + (q - r)2 ;
see Fig. 5.4. Our problem is to determine r so that PR + RQ is a
minimum, i.e., so that
y'a2 + r2 + y'b2 + (q
_
r)2
is a minimum.
p�
R
Figure 5.3. Path of a reflected ray of light
Figure 5.4. Determmation of the point R
We apply the triangle inequality as follows:
y'a2 + r2 +
y'b2 + (q - r)2 �
y'(a + b)2 + (r + q - r)2
= y'(a + b)2 + q2 .
Thus the distance traveled cannot be less than the fixed amount
y(a + b) 2 + q2 ; and this minimum value is achieved precisely if
the sets (a, r) and (b, q - r) are proportional with a positive constant
of proportionality, that is, if
(5.7)
�=
r
b_ >
_
0.
q- r
Observe what the condition (5.7) means geometrically. It means that
the right triangles ORP and TRQ are similar and that, since b is posi­
tive, q - r is positive so that R falls between 0 and T. From the
similarity of the right triangles, we conclude that the angles ORP and
TR Q are equal. Therefore, their complements 4 SRP and 4 SRQ
85
M A X I M I ZAT I 0N A N D M I N I M I Z ATI 0 N
are also equal. Thus Fermat's principle has enabled us to deduce the
famous result that the angle of incidence is equal to the angle of
reflection,
4 SRP
=
4 SR Q,
and the more obvious fact that R lies between 0 and T.
The above principle of reflection can also be proved by purely
geometric considerations. If R' in Fig. 5.5 denotes any point on the
x-axis different from R, and if Q' is the point (q, b) then
-
PR
+
RQ
=
PR
+
RQ'
=
PQ' < PR'
+
R'Q'
,
=
PR'
+
R'Q,
so that the point R is again seen to minimize the distance PR + RQ .
y
p
Q'
Figure 5.5. Geometric determination of the point R
Suppose a plane divides two homogeneous media M1 and M2 of
different density so that in M1 light rays travel with velocity v 1
along straight lines and in M2 they travel with velocity v2 along
straight lines. We now wish to determine the least time-consuming
path from a point P in M1 to a point Q in M2 ; see Fig. 5.6. Again
we have
PR
y
P
=
ya2
+
r2 ,
RQ
=
yb2
+
(q - r)2
(o, a )
Figure 5.6. A refracted ray of light
86
A N I NT R O D U CTION TO I N E Q U ALITI ES
and wish to minimize the time (distance/velocity)
yb2 + (q - r)2
.
v2
v1
While this is an easy problem in differential calculus, the minimum
value does not seem to be obtainable directly from our elementary
inequalities. The minimum value of t is obtained if the path from P
to Q is a broken straight line PRQ (see Fig. 5.6) located so that the
angles fh and fJ2 (between PR and the normal to the plane and RQ
and the normal to the plane, respectively) satisfy the relation
1
=
...ja2 + r2
+
sinfJ1
sinfJ2
This relation is known
as
v1
v2
Snell's law of refraction.
5.6 Simplified Three-dimensional Version of Dido's Problem
Consider now the problem of determining the rectangular box that
encloses the greatest volume for a fixed surface area; see Fig. 5.7.
Designating the lengths of the sides by x, y, and z, we see that the
volume is given by the expression
V = xyz ,
while the surface area is represented by the formula
A = 2xy + 2xz + 2yz .
y
Figure 5.7. Three-d1mensional version of the s1mpltfied problem of Dido
Given the value of A, we wish to choose the values of x, y, and z
that make V as large as possible. Once again, an application of the
arithmetic-mean-geometric-mean inequality will resolve the problem
for us.
Considering xy, xz, and yz as three nonnegative quantities, we
obtain the inequality
(5.8)
xy
;
+ x + yz ;::: [(xy)(xz)(yz)) l13
=
(xyz)213 .
M A X I M I Z AT l 0 N A N D M I N I M I Z AT I 0 N
87
We know that equality can occur if and only if
xy = xz = yz ,
a relationship that holds i f and only if
x =y = z.
Since
A
xy + xz + yz = 2 ,
the inequality (5.8) yields
1
2
2
� (xyz) 13 = V 13
or
(A)312
6
� v.
It follows that the volume of the rectangular box with surface area
A is less than or equal to (A /6)31 2 , and that this value is attained if
and only if
X =y = Z =
(�)1 12 •
Hence, the rectangular box of maximum volume for a given surface
area is a cube; and, dually, the rectangular box of minimum surface
area that encloses a given volume is a cube. Once again, we observe
a reciprocal relationship between two problems.
Exercises
I . Show that, if the sum of the lengths of the twelve edges of a rectangular
box is a fixed value E, then the surface area A of the box is at most
£ 2/24 , and that the box is a cube if and only if £ 2/24 = A .
2. State and prove the reverse of the result expressed in Exercise I.
3. Of all rectangles having the same length of diagonal, determine which has
the greatest perimeter and determine which has the greatest area. (Use
the results in Exercise 3 on page 24 or Exercise 3 on page 62.)
5.7 Triangles of Maximum Area for a Fixed Perimeter
Let us now consider the problem of determining the triangle of
maximum area for a given perimeter. In Fig. 5.8, let s denote half
AN I N T R O DU CTION TO I N E Q U A L I T I ES
88
Y
�
X
z
Figure 5 8 The problem of Dido for a triangle
the length of the perimeter of the triangle shown; that is, let
_
s - x +y + z.
2
As is well known, the area A of the triangle can be expressed by the
formula
A = [s(s - x)(s - y)(s - z)) 112 .
We wish to find the maximum value of the area as x, y, and z vary
over all positive values for which the relationship
2s = x + y + z
holds, where the value s is fixed.
Once again, the arithmetic-mean-geometric-mean inequality yields
the solution in a very simple way. For the three nonnegative values
s - x , s - y, and s - z , we have
(s - x) + (s
[(s - x)(s - y)(s - z)) 1 13 :s;
<
-
3s - (x + y + z) = 3s - 2s = � .
3
3
3
Hence,
(5.9)
� y) + (s - z)
(s - x)(s - y)(s - z) :s;
GY·
From (5.9), by easy steps, we obtain the result
A = [s(s - x)(s - y)(s
, z)]112 [s(§..)3] 112 = ( 543)1 12 =
_:
:s;
3
There is equality if and only if
3
s2
3\73 "
s - x = s - y = s - z,
that is, i f and only i f x = y = z . Consequently, we can assert:
M A X I M I Z A T I 0 N A N D M I N I M I Z AT I 0 N
89
THEOREM 5. 1 . Of all triangles with a fixedperimeter, the equilateral
triangle maximizes the area.
Observe that all the results we have so far obtained in this chapter
seem to indicate that symmetry and optimal behavior are closely in­
terrelated. Perhaps the profound and aesthetic observation of Keats,
"Beauty is truth, truth beauty," sums this up best.
Exercises
1 . Let s denote half the length of the perimeter and A the area of the triangle
having sides of length x, y, and z, so that
s - x +y + z
2
and
A2 = s(s - x)(s - y)(s - z) .
For the special case of an isosceles triangle, with y = z , let the area be
denoted by /; and for the special case of an equilateral triangle, with
x = y = z , let the area be denoted by E. Show that
/2 =
and
� x2(s - x)
2. With the notation of Exercise 1 , show that, for triangles of equal perimeter s,
and
J2 -
{
A2 = <s - x)(y - z)2 ,
where x is the length of the base of the isosceles triangle and also the length
of one side of the general triangle.
3. Using the formulas in Exercise 2, show that
£ 2 - [ 2 2': 0
and
J 2 - A 2 2': 0 .
Under what circumstances do the signs of equality hold?
4. Using one of the inequalities in Exercise 3, show that, if the perimeter and
one side of a triangle are given, then the area is maximized by making the
triangle isosceles.
5. Using one of the inequalities in Exercise 3, show that of all isosceles tn­
angles with a given perimeter, the equilateral triangle maximizes the area.
90
AN INTRO D UCTION TO I N EQU ALITIES
6. Using the inequalities in Exercise 3, explain how the formula
A = y£2
_
(£2
_
J 2)
_
(12
_
A2)
exhibits A in terms of nonnegative expressions, and how the results stated
in Exercise 4, Exercise 5, and Theorem 5. 1 follow from this formula.
7. Of all right triangles having the same length of hypotenuse, determine
which has the greatest altitude to the hypotenuse.
5.8 The Wealthy Football Player
Consider now a problem of a similar but slightly more complex
type.
A football player, having parlayed his athletic prowess on the
gridiron into a position in Wall Street and thence by natural stages
into the possession of a variety of stocks and bonds of his own, be­
came, in the course of time, quite wealthy. Being of a sentimental
turn of mind, as retired football players are wont, he stipulated in
his last will and testament that he be buried in a vault shaped like a
giant football. His executors, honoring his last request, were faced
with the problem of carrying out this assignment in the most eco­
nomical fashion.
After some thought, they decided that the mathematical problem
most closely approximating the physical situation that they faced was
that of enclosing a rectangular box ofgiven dimensions within an ellips­
oid of the least possible volume.
Following the leads contained in the foregoing problems, they
began with the reverse problem, i.e., that of inscribing a rectangular
box ofgreatest volume in a given ellipsoid
According to solid analytic geometry, an ellipsoid with center at
the origin and axes along the coordinate axes has an equation of the
form
(5. 10)
where 2a, 2b, 2c represent the lengths of the axes of the ellipsoid. Now,
it is intuitively clear that the inscribed rectangular box will have its
center at the origin and its sides parallel to the axes. Thus, if one ver­
tex of the box is at the point (x, y, z) on the surface of the ellipsoid
(see Fig. 5.9 which shows one eighth of the ellipsoid), then the other
seven vertices must be located symmetrically. This means that the
M A X I M I Z AT I O N A N D M I N I M I Z AT I O N
91
z
( o,o,c )
X
Figure 5 9. One octant of a box in an ellipsoid
coordinates of the other seven vertices of the box are given by
( - x, y, z) ,
(x, -y, z) ,
(x, y, - z ) ,
( - x, -y, z) ,
(x, -y, - z) ,
( - x, y, - z) ,
( -x, -y, -z) .
Since the sides of the box have lengths 2x, 2y, 2z, it follows that
the volume of the box is given by the expression
(5. 1 1 )
V = 8xyz .
The problem that confronts us is that of maximizing the expression
(5. 1 1 ) subject to the condition that the quantities x, y, and z satisfy
eq. (5. 10).
Once again, this problem is easily resolved by use of the arithmetic­
mean-geometric-mean inequality. We have
(5. 12)
Using the fact that x, y, and z satisfy eq. (5. 10), and using the formula
(5. 1 1 ) for V, we see that the inequality (5. 1 2) yields
.l >
v 213
3 - 4(a2b2c2) 113
or
1 <abc)213 ;::: V 213 .
It follows that the volume of the box is at most Sabc/3 y'3; further­
more, by the condition for equality in (5. 1 2), this value is attained if
92
AN I NTR0 D UCTI0N T0 I N EQ U AL JTIES
and only if
or
a
X = \73 '
(5. 13)
b
y = \73 '
c .
= \73
z
These values (5. 1 3) furnish the desired dimensions of the box,
given an ellipsoid with semiaxes a, b, and c.
Now let us return to the original problem. We have seen that for
given values a, b, c, the values (5. 1 3 ) for x, y, z give the lengths of
the half sides of a rectangular box of maximum volume. But in the
other direction, is it true that for given values x, y, z, the values
c = -Jj z
a = -Jj x,
(5. 14)
give the minimum volume for a containing ellipsoid? Whether the
executors were perspicacious or just lucky, they were right in follow­
ing their mathematical hunch. Here is a proof of that fact:
The volume W of the ellipsoid (5. 10) is given by the formula
4 abc,
W = 3'1T
and for given positive x, y, z we want to choose positive a, b, c satis­
fying eq. (5. 10) in such a way as to minimize W.
In order to see the reciprocal relationship between the two prob­
lems, it is convenient to consider reciprocals; and since it is now
x, y, z that are given and a, b, c to be determined, we also reverse
ends of the alphabet and set
1
(5 . 1 5) a X'
-
b= I , c- _!_ ·
Y
Z'
I
I
X = A, y = B,
Z =
I
c
-
.
Eq. (5. 10) then becomes
( l /A)2
( l /B)2
( l / C )2
+
+
( l /X )2 ( l / Y)2 ( I /Z )2
=
I
or
(5. IO')
and, subject to this constraint, we wish to choose X, Y, Z so as to
M A X I M I Z AT I O N A N D M I N I M I Z A T I O N
93
minimize
W = 34 ?T abc .
Since the coefficient 4?T/3 is a constant, minimizing W is equivalent
to minimizing
abc =
1
XYZ
which, in turn, is equivalent to maximizing
V ' = 8XYZ .
(5. 1 1 ')
But this is just the problem we have already solved, namely, the
problem of determining the box of maximum volume that can be in­
scribed in a given ellipsoid. It does not matter that this ellipsoid
(5. 10') and box of volume (5. 1 1 ') have no physical existence relative
to our football player; indeed, this fact serves to emphasize and
dramatize the importance of pure mathematical analysis. The solu­
tion values [cf. eq. (5. 1 3)] are
(5. 13')
A '
X = -Jj
y=
B '
-Jj
Substituting from (5. 15) into (5. 13'), we obtain (5. 13) and (5. 14).
Considering, in a typical case, a length 2x of 6 feet, a width 2y of
2 feet, and a height 2z of I foot, we obtain the values
a=3
-Jj ,
c = yJ/2 ,
for an ellipsoid of minimum volume that would contain the box.
Because of the curvature of the elliptical vault, there would be
ample room for the inclusion of a few well-worn footballs to accom­
pany our gridiron hero to his place of rest.
5.9 Tangents
Let us now apply the theory of inequalities to the problem of find­
ing a tangent to a given curve. Of course, we can do this only in an
intuitive fashion at this time, since any precise definition of what we
mean by a tangent to a curve at a point lies outside our chosen
domain of discourse.
94
AN I NT R O D U CT I ON TO I N E Q U A LITIES
Consider a curve determined by an equation of the form y
as shown in Fig. 5. 10, and a line, with equation
(5. 16)
mx + ny
=
= f(x) ,
k,
that cuts the curve in the points P and Q. Let us move the line parallel
to itself [this merely involves changing the value of k in the equation
(5. 1 6) of our line] until these two points become coincident at R, as
shown in Fig. 5. 1 1 . The line mx + ny = k' may be called the tan­
gent line to the curve y = f(x) at the point R. Let us again point
out that we are proceeding in an intuitive fashion without attempt­
ing to make the notion of a tangent precise. But you are familiar at
least with tangents to circles and can see that the procedure does lead
to the desired result in this special case.
y
y
Figure 5 10 A curve and a cutting lme
0
Figure 5. 1 1 . A curve and a tangent line
We shall use the foregoing procedure and the theory of inequalities
in order to determine the tangent lines in a given direction to an
ellipse. Suppose the ellipse (see Fig. 5 . 1 2) is given by the equation
x2 y2 _ 1 ,
a2 + b2 and let (x 1oy 1) be one of the two points of tangency for lines of the
form mx + ny = k, where m and n are fixed, m2 + n2 -=!= 0 , and k
varies through all values. Observe that this point of tangency can be
characterized in the following way:
(a) The point (x1 , y 1) lies on the ellipse; i.e., the values x1, y1
satisfy the equation
X12 Y 12 - 1 .
(5. 17)
a2 + v (b) The point (x 1 oy 1) lies on the line; i.e., the values x1o y1 satisfy
the equation
mx 1 + ny1 = k .
M A X I M I ZAT I O N A N D M I N I M I Z AT I O N
95
y
Figure 5 . 1 2 An ellipse and two parallel tangent lines
(c) The distance from the origin to the line mx + ny = k is a
maximum for all points (x 1 .y 1) satisfying conditions (a) and (b), as
k varies.
The distance d from the origin to the line mx + ny = k is given
by the formula
lkl
! mx 1 + ny1 i .
d(5. 1 8)
- ym2
2
ym 2 + n2
+ n
To see this, observe that the line
(5. 19)
mx + ny
=
k
has slope - min (see Fig. 5.13) and that a perpendicular line OP
through the origin must therefore have the equation
n .
y = -x
m
(5.20)
Solving the system of linear equations (5. 19) and (5.20), we find the
y
Figure 5 13. The distance formula
96
AN I N T R O D U CTION TO I N EQ U A LI T I ES
coordinates of P to be
x=
km
,
m2 + n2
y=
m2
kn
+
n2 ·
The distance from the origin to P is therefore
[ k2m2 _...:.k;:_2...:.n_2-J 112
(m2 n2 )2 (m2 n2)2
_ ( k2 )v 2 _
1k1
- m2
2
2
n2
d = OP =
+
+
+
+
'Jm + n '
and, since k = mx1 + ny1 by condition (b), formula (5. 1 8) is proved.
It follows that the problem of determining the tangent line is that
of maximizing the expression (5. 1 8) over all x1 and y1, subject to the
constraint (5. 1 7).
An application of Cauchy's inequality, i.e., the inequality (4.38) in
Sec. 4.4(a) of Chapter 4, yields
_ mx + ny
(am)(x 1 /a) + (bn)(y1 /b) l
d - 1 1 2 + 21 l - l
n
ym
ym2 + n2
_
(5.21)
The points of tangency are determined by two conditions. Namely,
they must lie on the ellipse, so that (x 1, y 1 ) satisfies (5. 1 7) : and the
distance (5. 1 8) must attain its maximum value, so that the equality
sign holds in (5.2 1 ), which is the case if and only if
(5.22)
The solution of the system of linear equations (5. 1 7) and (5.22) yields
the values
(5.23)
_ -+- -...:b:...
.. n
: 2.:.:
y1 - ..,...2 ....,. 2 + b:-:---2n2)112 '
(a m
where either both plus signs or both minus signs are taken, and the
desired values k' and k" (see Fig. 5 . 1 2) of the constant k can then be
found by substitution in eq. (5. 19).
M A X I M I Z AT I O N A N D M I N I M I Z AT I O N
97
5.10 Tangents (Concluded)
With a slight amount of additional effort, we can obtain a much
more elegant result, resolving the problem of finding the tangent to
an ellipse at a given point on the ellipse, rather than the tangent in a
given direction.
In place of solving for x 1 and Yt in terms of m and n, we need only to
solve for m and n in terms of X t and Yt · It is that simple! From
eq. (5.22), we see that we must have
n = !l.!
b2
where r is some as-yet-undetermined constant of proportionality.
Using these values in the equation mx + ny = k, we obtain the
equation of the tangent line in the form
or
(�t) x + ('�t)y
=
k
Since, on the one hand, (x 1, y 1) is a point on this tangent line, and
since, on the other hand, it is also on the ellipse-that is, (xt.y 1) sat­
isfies equation (5. I 7)-we must have klr = I . Hence, we obtain the
very simple and elegant result that the tangent to the ellipse
x2 y2
I
+
a2 b2
-
_
at the point (Xt .Yt) on the ellipse has the form
XXt + ll.!_ = I .
b2
a2
As you will see when you study calculus, this is the same result that
one obtains through the use of derivatives.
Exercises
I . Determine which of the points (5, - 3), (3, 5), and (7, 0) is on the ellipse
x2
50
+
y2
_
18 -
I.
98
A N I NT R O D U CTION TO I N EQUALITI ES
2. For x = 2 , determine the values y such that the point (x, y) lies on the
ellipse in Exercise 1. Also for x = 2 , detenrune the values ofy such that
the point (x, y) hes either inside or on the ellipse.
3. Determine the equation of the tangent to the ellipse in Exercise 1, at the
point ( - 5, 3) on the ellipse.
4. Solve the system of equations
x2 y2
+
1,
a 2 b2 _
mx + ny = k ,
where a, b, m, and n are fixed, choosing k so that the resulting quadratic
equation has a double root. (You will find that k2 = a 2m 2 + b2n2 .)
Compare your answer with the values of Xt, Yt given at the end of Sec.
5.9, page 96.
CHAPTER SIX
Properties of Distance
6.1 Euclidean Distance
The distance with which you are familiar, between two points
P : (xt.Jt) and Q : (x2 ,y2) in the (x, y)-plane, is called Euclidean dis­
tance. We shall denote it by d2(PQ); it is given by the formula
(6. 1)
d2(PQ)
=
[(x2 - Xt)2
+
(y2 - Yt)2] 11 2 .
We shall now enumerate some of the properties that characterize
this distance function.
I. The distance between two points depends only on the position
of one relative to the other; that is, it depends only on the differences
x 2 - Xt and Y2 - Yt of their coordinates. This property (that the
distance between two points does not change when both points are
shifted by an equal amount in the same direction) is called translation
invariance.
2. The distance from a point P to a point Q is equal to the distance
from Q to P. This is seen by venfying, in (6. 1 ), that
Property 2 is usually called the symmetry of the distance function.
99
100
A N I NTRO D UCTION TO I N EQUALITIES
3.
The
triangle inequality
d2(PR) ::; d2(PQ) + d2(QR)
is satisfied by the distance function (6. 1); see Sec. 4.6.
4. The distance d2(PQ) between any two points is
for all P and Q. That is,
nonnegative
d2(PQ) z 0 ;
the sign of equality holds if and only if the points P and Q coincide.
This is often called the positivity of the distance function; it follows
immediately from the definition
a
5.
(6. 1).
Q has
If P has coordinates (x,y) and
is a nonnegative constant, then
(ax, ay), where
d2 (0Q) = ad2(0P) ;
origin (0, 0). This property is
0 denotes the
sometimes called
homogeneity of the distance function, and it holds because
d2(0Q) = [(ax)2 + (ay)2) 11 2 = [a2(x2 + y2)) 112
= a(x2 + y2 )1 12 = a d (0P) .
2
here
the
coordinates
The Euclidean distance has still another property:
6.
The Euclidean distance between two points remains unchanged
if the (x, y)-plane is rotated about the origin through some angle.
This property is sometimes called rotation invariance.
6.2 City-Block Distance
It turns out that many other useful and interesting "non-Euclidean"
distances can be defined. In order to be called a "distance," a func­
tion of P and Q must have the properties 1 through 5 that we just
verified for the familiar distance
alone has all six properties.
( 6. 1 ).
The Euclidean distance
d2
As an example, let us invent a "city-block distance" by determining
the actual length of a path between two addresses
Q : (x2 , y2 )
P : (x 1 .y 1)
and
in your home town, assuming that all the streets are laid
out strictly north-south and east-west, and that there are no empty
lots to cut across; see Fig. 6. 1 . Any path from P to Q is made up ex­
clusively of horizontal and vertical pieces, so that the distance d(PQ)
we would have to traverse consists of the sum of all the horizontal
distances and all the vertical distances making up the path from P to
IOI
P RO P ERTI ES O F D I STAN C E
y
STREETS
6
5
4
3
2
0
p
Q
2
3
AV E N U E S
4
5
6
X
Figure 6. 1 . City-block distance
Q. We shall therefore define city-block distance d1 by
(6.2)
although, strictly speaking, this does not quite accurately describe
the situation. It is not correct for the case shown in Fig. 6.2, where
the addresses P and Q lie between the same two north-south (or
east-west) streets; in this case, the traveler is forced to reverse direc­
tions in the course of his walk. Nevertheless, let us take (6.2) as the
definition of a new "non-Euclidean" distance. Mter all, our city-block
example just served as a motivation. (If our blocks are very small,
(6.2) is fairly accurate. More precisely, our new distance function is
the minimum distance required in traveling from P to Q when con­
strained to move only in the four principal compass directions.)
Figure 6.2. City-block distance, exceptional case
Next, let us see if d1. as defined by (6.2), has the five properties
required of a distance.
Since only the difference of the coordinates enters the expression
(6.2), our new distance is certainly translation invariant, so that it has
property 1 .
Since d1(PQ) = d1(QP ) , d1 i s symmetric, and thus i t has prop­
erty 2.
102
A N I NT R O D U CTION TO I N EQ U A L I T I ES
To establish the triangle inequality
d1(PR) :::; d1(PQ) + d1(QR) ,
let P, Q, R have coordinates (x 1 ,y 1), (x2 ,y2), and (x3, y3), and write
l x1 - x3 1 + IY1 - Y3 1
= I X ! - X 2 + X2 - X3 l + IYl - Y2 + Y2 - Y3 l ;
since, by Theorem 3.2, page 42,
d1(PR)
=
I x 1 - x2 + x2 - X3 I :::; I x1 - x2 l + I x2 - x3 1 ,
IY 1 - Y2 + Y2 - Y3 1 :::; IY 1 - Y2 l + IY2 - Y3 1 ,
we have
d1(PR) :::; l x1 - X2 l + IY1 - Y2 l + l x2 - X3 l + IY2 - Y3 1
= d1 (PQ) + d1(QR) .
The city-block distance certainly satisfies our fourth condition
since the absolute value of any real number is always nonnegative.
It is positive unless P and Q coincide.
Property 5 is easily verified since, for a 2 0 ,
I ax I + I ay I = a[ I x I + IY I ] .
Next, let us try to generalize the notion of a circle from Euclidean
to city-block geometry. In Euclidean geometry, a circle is the locus
of points equidistant from a fixed point. Let us carry this definition
over to our new geometry. According to (6.2), the "unit circle" with
center at the origin 0 : (0, 0) would be given by the equation
l x l + IY I = 1 .
Its graph in the ordinary Euclidean plane is shown in Fig. 6.3 (see
also Fig. 3. 1 4).
d1(0P)
=
y
Figure 6 3 The unit circle in city-block geometry
P RO P ER T I E S O F D I STAN C E
103
6.3 Some Other Non-Euclidean Distances
Suppose that we now define the "distance" between the origin 0
and an arbitrary point P to be given by the expression
(6.3)
for some fixed value of p 2 I , and generally, that we define the dis­
tance between any two points P : (xt.y 1 ) and Q : (x 2 ,y2) to be
given by
(6.4)
dp(PQ) = [ I Xt - X 2 1 p + IY t - Y2 1 P) l!p .
Again, we check the five distance properties. The distance (6.4)
certainly is translation invariant. Moreover, it is symmetric; that is,
dp(PQ) = dp(QP) . Thirdly, the triangle inequality follows from for­
mula (4.53'), which was proved at the end of Sec. 4.8 from the Min­
kowski inequality. Property 4 on positivity is also satisfied by (6 .4);
and finally, for P: (x, y) and Q : (ax, ay) , we have
dp(OQ) = [ l ax i P + l ay i P)l iP = (aP )liP( I x i P + IY I P) 11P = a dp(OP) ,
so that our distance dp also possesses the fifth property.
In this case the "unit circle," i.e., the locus of points at a distance
I from the origin, is given by
I X I p + IY I p = I .
Just what such locus curves look like depends on the particular value
ofp. For example, when p = I , we are back at the city-block unit
Figure 6 4 Euclidean graphs of non-Euchdean "unit circles"
104
A N I NT R O D U CTION TO I N EQ U A LITI ES
circle; for p = 2 , we have the usual Euclidean unit circle. For
p = 1 , 2, 4 , we have the inequalities
i x i + IY I � ( i x i 2 + IY I 2) 11 2 � ( i x l 4 + IY I 4) 114 •
which are easily verified by squaring. The Euclidean graphs of the
corresponding "unit circles" are shown in Fig. 6. 4; the curve for
p = 1 is contained in that for p = 4 . Is it true, in general, that the
"unit circles" for distances defined by (6.3) are situated in such a way
that the one for any fixed p contains those corresponding to smaller
p? If so, do these curves become larger and larger as p increases?
To answer the first question, we first formulate an equivalent
question: Does the inequality
[ ! x i " + IY I ") l !n � [ i x l m + ly l m) l!m
hold whenever m � n � 1 ? It does indeed. Let us give a proof for
m = 3 , n = 2 , and leave the general case for the reader as an
exercise.
In order to avoid writing so many absolute-value signs, let us
introduce
a = ix l ,
b = IY I ·
We must then prove
(a2 + b2) 1 12 � (a3 + b3) 11 3
for a, b � 0 .
We write
a3 + b3 = aa2 + bb2
and apply Cauchy's inequality (in square-root form), obtaining
(6.5)
Since
(6.5) yields
or
whence
The proof for arbitrary rational m � n � 1 can be obtained by a
corresponding application of Holder's inequality. It is not difficult to
P R O P E R T I E S O F D I S TA N C E
105
see from the above discussion that, as p approaches 1, the "unit
circles"
I X I P + Iy i P = 1 ,
p> 1,
approach the square I x I + IY I = 1 shown in Fig. 6.3.
In order to see what happens as p gets larger and larger, we observe
that
(6.6)
max { 1 x 1 v, ly l v} :::; 1 x 1 v + ly l v :::; 2 max { 1 x 1 v, ly i P } .
Since
[max { I x I v, IY I P } ) 1 1P = max { I x I , IY I } ,
(6.6) yields
(6.7)
max { l x i . IY I } :::; [ l x i P + ly l v) llv :::; 21 1v max { l x i , IY I } .
Now consider what happens to the right-hand member of (6.7) as p
becomes very large; p occurs only in the exponent l ip of 2. As p be­
comes very large, l ip becomes very small and 2 11P therefore ap­
proaches 20 = 1 . In other words, (6.7) tells us that
dp(OP) = ( l x i P + ly i P) 11P
approaches the distance
(6.8)
d"'(0P) = max { I x I , ly I }
as p "approaches infinity." It can be shown that d"' has all five
distance properties.
Now, what does the "unit circle"
max { I x I , IY I } = 1
look like? It is given by the square with sides
(6.9)
lxl = 1 ,
IY I = 1 ,
0 :::; IY I :::; 1
0 :::; l x l :::; 1 .
So we see that
1
dv(PQ) = [ l x1 - x 2 I P + IY 1 - Y2 I P] 1P
can be used as a distance function for all p 2 1 and that, asp grows
larger, the "unit circles"
I X I P + IJ I P = 1
grow larger, approaching the square (6.9) as p approaches infinity;
106
A N I NTR0D U CTI 0N T0 IN EQ UALITIES
in spite of the fact that these "unit circles" grow bigger and bigger,
they never get out of the square (see Fig . 6.4).
The "unit circle," in all these cases, divides the (x, y)-plane into two
regions : the interior of the "unit circle," comprised of all points at dis­
tance less than 1 from the origin, and the exterior, comprised of all
points at distance greater than 1 from the origin. The set of points
defined by the inequality
d( OP) ::; 1
is sometimes called the unit disc, and the "unit circle"
d(OP) = 1
is called the boundary of the unit disc .
Some general remarks are in order . The Euclidean distance, as
pointed out earlier, is invariant under translations (property 1 ) and
rotations (property 6), i.e., under the so-called displacements or
rigid-body motions. The other distances treated above also do not
change under translation, but they do change under rotation. In fact,
we can see from Fig. 6.4 that the city-bloc k distance d1 goes into the
distance d"' = max { I x2 - Xt l , IY2 - Yt l } (stretched by a factor
-J2) when the (x, y)-plane is rotated through 45° about the point
(x1 , y 1 ). It can be shown (but we shall not do so here) that the
Euclidean distance is completely characterized by the six properties
enumerated in Sec . 6. 1 ; this means that the only distance that has, in
addition to the five required properties, the property of rotation
invariance is the Euclidean distance.
6.4 Unit Discs
There are m any other functions having the five properties required
of a distance ; we have considered only a few.
We might ask the following question : Given a set S of points
(which we shall henceforth call a point set) in the (x, y)-plane, such that
S contains the origin in its interior, under what conditions does it
represent the unit disc belonging to some distance d? In other words,
under what conditions does there exist a distance function dfor which
S contains exactly those points P that are characterized by the
inequality
d( OP) ::; 1 ?
P RO P E R T I E S OF DI STAN C E
107
We claim that, for a given point set S having the origin in its interior
there exists a distance function d such that S plus its boundary is the
unit disc for d if and only if the following conditions are satisfied:
(a) The point set S is symmetric with respect to the origin.
(b) The point set S is convex.
A point set S is symmetric with respect to the origin if, for each
point (x, y) belonging to S, the point ( - x, -y) also belongs to S. It is
convex if, for each pair of points in S, the line segment joining these
points is entirely in S. See Fig. 6.5, (a) and (b).
y
y
(a)
(b)
Figure 6.5. Point sets i n the plane
(a) Pomt set symmetric with respect to the origin
(b) Convex Point set
We shall first prove that property (a) holds if d is a distance func­
tion: If d is a distance function and S its unit disc, then S is symmetric
with respect to the origin. In other words, if d is a distance function
and P: (x, y) satisfies d( OP) s I , then the point Q : ( - x, -y)
satisfies d(OQ) s 1 .
If d is a distance, then it is translation invariant; hence, if we shift
the points Q and 0 by an amount x in the horizontal direction and
an amount y in the vertical, their distance d( QO) will remain un­
changed. But such a shift brings the point Q into the origin 0 and
the origin 0 into the point P. Hence
d( QO)
=
d( OP) S I,
=
d( QO) s 1 .
and, since d is also symmetric,
d( OQ)
Accordingly Q lies in the unit disc.
108
A N I N T R O D U CTION TO I N EQ U A L I T I E S
Next, let us prove that if d is a distance function then property (b)
holds by showing that, if P and Q are any two points in the unit disc,
then the line segment PQ lies entirely in the unit disc. Symbolically,
the problem is expressed thus: Given any two points P, Q such that
d(OP) s I , d(OQ) s I for some distance function d, and any given
point R on the segment PQ, prove that d(OR) s I . This is imme­
diately seen to be true if P = Q or if either P or Q is at the origin,
so we may assume that 0, P, and Q are distinct points.
y
0
Figure 6.6 Convexity of the unit disc
The first step in the proof consists in expressing the fact that R lies
on PQ in a convenient form. Let P'R be parallel to OQ and let Q'R
be parallel to OP (see Fig. 6.6). Then we claim that
(6. 10)
d(OP')
=
a d(OP) ,
d(OQ')
=
b d(OQ) ,
where a > 0 , b > 0 , and a + b = 1 . To see this, we shall first
employ the Euclidean theory of proportions (similar triangles). Using
the same symbol AB for a Euclidean line segment and its length, we
have
(6. 1 1)
where a and b simply denote these ratios and hence are positive
numbers. Moreover, by adding these ratios, we obtain
a+b
=
QR +
RP
QP
=
QP
QP
=
I.
Since P and P' lie along the same straight line through 0, their
coordinates are proportional. We may therefore make use of prop­
erty 5 of the Euclidean distance and deduce, from ( 6 . 1 I ), that if P
has coordinates (X t. Yl), then P' has coordinates (ax1. ay1). Similarly,
109
P R O P E R T I E S O F D I S TA N C E
since Q and Q' are on the same line through 0, if Q has coordinates
(x2,y2), then Q' has. coordinates (bx2 , by2). We now use the fact that
our distance d also has property 5, so that
d(OP')
(6. 12)
=
d(OQ')
a d(OP) ,
=
b d(OQ) .
Finally, we use property 3 (the triangle inequality; see Fig. 6.6):
or, since d(P'R)
=
d(OR) S d(OP') + d(P'R) ,
d(OQ') by property 1 ,
d(OR) S d(OP') + d( OQ') .
Substituting from (6. 1 2), we have
d(OR) S a d(OP) + b d(OQ) ,
and, since d(OP) s 1 , d(OQ) s I , and a + b
d(OR) s a + b
=
=
I,
I,
which, by definition, means that R is in the unit disc.
So far we have shown that, if d is a distance and S its unit disc,
then S is symmetric with respect to the origin and convex. We must
still prove the converse: If S is a point set containing the origin in its
interior, and S is convex and symmetric with respect to the origin, then
there exists a distance function dfor which S is the unit disc.
We shall indicate how such a distance d may be defined, but we
shall leave to the reader the job of verifying that d has the five pre­
scribed distance properties.
y'�
y
z
/ 1' ,
I_
Q
\a·-----­
- + -f-1==
- - - x'
I
\\ I I
I
'r
Figure 6 7. d( OZ)
=
OZ ,
OZ'
d(PQ)
=
!!2_
PQ'
1 10
A N I N T R O D U CTION TO I N EQ U A L I T I ES
Let S be a point set of the prescribed sort; see Fig. 6.7. Let Z be
any point other than 0 in the plane. Draw the ray from 0 through
Z and let it intersect the boundary of S in the point Z'. (It follows
from the convexity of S that there is just one such intersection of the
ray with the boundary of S.) Then calculate the ratio
r = oz
OZ'
of the Euclidean distances OZ to OZ' and define the distance from
0 to Z to be this ratio; i.e., let
d(OZ) = r .
Observe that d( OZ) is less than 1 , equal to 1 , or greater than I de­
pending on whether Z is an interior point of S, a boundary point of
S, or an exterior point of S, respectively.
To define d(PQ) for any points P and Q, shift the coordinates as
indicated in Fig. 6.7 and proceed as before.
In an ordinary Euclidean circle, the ratio of the circumference to
the diameter is denoted by the symbol ., and is approximately 3. 14.
In the exercises below, the task is to find the ratio r of the non­
Euclidean length of the circumference to that of the diameter of some
non-Euclidean unit circles. We shall analyze the situation in one par­
ticular case, below, and leave other simple cases for the reader to
investigate.
y
Figure 6.8. A regular hexagon symmetric With respect
to the co-ordinate axes
ExAMPLE. Let S consist of the points of a regular hexagon symmetrically
situated with respect to the coordinate axes, together with the points of its
interior; see Fig. 6.8. Since S is convex and symmetric with respect to the
origin, it may be considered the unit disc for some distance function d. Since
P R O P E R T I E S O F D I S TA N C E
Ill
the non-Euclidean radius of the unit disc is I by definition, its diameter is 2.
To calculate the circumference, we observe that, because of the translation
invariance of d, the following relations hold:
d(P1P2) = d(OP3) = I , d(P2P3) = d(OP4) = I , d(P3P4) = d(OPs) = I ,
d(P4Ps) = d(OPs) = I , d(PsPs) = d(OPt) = I , d(PsPt) = d(OP2) = I .
I f we add the lengths o f all these segments, we find that the circumference
is 6. Hence, the desired ratio is
r=
� = 3.
Exercises
Calculate the ratio of the non-Euclidean length of the circumference to that
of the diameter for the following unit discs S:
I. S is the unit disc for the city-block distance I x2 - Xt l + IY2 - Yt l ; see
Fig. 6.3.
2. S is the unit disc for the distance function
d(0P) = max { I x I , IY I } .
3. The unit disc S consists of the interior and boundary points of a regular
octagon with center at the origin.
4. The unit disc S consists of the interior and boundary points of a regular
ten-sided polygon with center at the origin.
6.5 Algebra and Geometry
What we have observed in the preceding sections is that geometric
intuition can be used to derive interesting algebraic results. In two
or three dimensions, this technique works well. As soon as we tum
to a discussion of n-dimensional geometry, for n 2 4 , the situation
reverses. Now, we often rely on algebra to make geometric definitions
and to establish geometric results.
Let us briefly illustrate this idea. Consider a set of n real quantities
X t. x2 , . . . , Xn as constituting a point P in n-dimensional space. The
Euclidean distance between the two points P : (xt. x 2 , . . . , Xn) and
Q : (y l , y2, . . . , Yn) is defined to be
(6. 13)
d(PQ) = [(x1 - y1)2 + (x2 - y2)2 + · · · + (xn - Yn)2) 11 2 .
For n = 2 , (6. 13) reduces to the familiar expression for the distance
between two points (xt. x2) and (yt.y2) in the plane. If we denote the
origin (0, 0, . . . , 0) by 0 and the point (x1 + Y t. x2 + y2 , . . . , Xn + Yn)
112
A N I N T R 0 D U C TI 0N T0 I N E Q U A L I T I E S
by R, then the n-dimensional triangle inequality
d(OP) + d(PR) ;::: d(OR)
would read
[x12 + x2 2 + . . . + Xn2] 1 12 + [y1 2 + y2 2 + . . . + Yn2]112
;::: [(xl + Yt)2 + (x2 + Y2)2 + · · · + (Xn + Yn)2] 112 .
This is a valid inequality, as was noted on page 70 in Sec. 4.6.
Next, let us define the cosine of the angle () between the lines OP
and OQ by
(6. 14)
(6 15) COS ()
+ Xo/'n
XJ..Yt + Xg,Y2 +
(x12 + x2 2 + . . . + Xn2)1 12(y12 + y22 + . . . + Yn2) 112
The Cauchy inequality [see Sec. 4.4(d), inequality (4.45)] shows that
1 cos 8 1 � 1 .
We now have the foundations for an analytic geometry of n
dimensions.
·
-
-
• • •
·
SYMBOLS
IaI
N
0
p
E
f
=I=
>
*
<
{ah a2 ,
max {a1. a2 ,
min {a1 , a2,
{ah a2 ,
{a1 , a2,
1:
�
l
�
i
�
v
an}
an}
an}
' an} +
, an} ­
sgn x
0
o
0
0 0
o
o
0 0
0
o o
o
0
,
,
o ,
absolute value of a
set of all negative numbers
set of one element, namely, the number 0
set of all positive numbers
is a member of
is not a member of
equal to
not equal to
greater than
not greater than
less than
not less than
greater than or equal to
neither greater than nor equal to
less than or equal to
neither less than nor equal to
neither greater than nor less than
nonnegative square root
set of elements a1, a2 ,
, an
greatest element of {a1 , a2 ,
an}
least element of { a1. a2 ,
an}
max {0, a1 , a2 ,
an}
an}
min {0, a 1. a2 ,
0 for x = 0 , I for x > 0 , - I for x < 0
0
0
0
0
0
o
0
1 13
0
0
0 ,
0 ,
o o ,
0
0 ,
0
A nswers to Exercises
Chapter 1
Pages 1 2- 1 3
I.
-1 5
-3
-3 < -2 <
-I
-2
-
2
0
1 5 < -I < 3 .
./2
3 - 7T 7T- 3
'TT
<0<
'TT -
3
+
3 < V2< 2 < 3
2. (a) e (b) j (c) f (d) f (e) e (f) f ( g) e (h) f (i) f (j) t
3. (a) N (b) P (c) P (d) N (e) P (f) P (g) P (h) N (i) P (j ) 0
4. (a) < (b) > (c) > (d) < (e) > (f) >
(g) > (h) < (i) > (J) =
5. ( a) T (b) T (c) T (d) F (e) T (f) T (g) F (h) T (i) F (j) F
6. 2 , "' - 3 , - (3 - w)2 , al(c - b) , 0 , - yb2 - 4ac .
7. (a) :;:::: (b) � (c) � (d) > (e) < (f)
=
S. p > O , - n > O . so p - n > O , so p > n .
9. a = b
1 15
1 16
A N I NT R0 DUCTI 0N T0 I N EQUALITIES
10. True for
n
= 1 by hypothesis; induction step by AxiOm II.
1 1 . Numbers 1 and 2 were shown in text to be "positive," so 1 + 2 = 3 is
"positive" by Axiom II. Let a = 1 I 3 ; then 3a = 1 , so a is "positive"
since 3 and 1 are "positive." Then 2a = 2 / 3 is "positive" by Axiom II.
Chapter 2
Page 19
1 . Add a/2 < b/2 first to a/2 = a/2 , then to b/2 = b/2 .
2. Both inequalities are equivalent to (ad - bc)2 :2:: 0 .
3. Equivalent to (a - b)4 :2:: 0 .
4. For n = 2 , the theorem states: I f a1 :2:: a2 , then a1 :2:: a2 , which is
true. Suppose true for n and let a1 :2:: a2 , . . . , a,._1 :2:: a,. , a,. :2:: a,.+ 1 .
Then a1 :2:: a,. , a,. :2:: a,.+ 1 , so a1 :2:: a,.+ l · Sign of equality if and only
if a1 = a2 , . . . , a,._l = a,., and a,. = an+ l ·
Pages 2 1 -22
1 . Intermediate step:
2 <!+!
ya!j - a b "
Equality if and only if a = b .
2. In Exercise 1 , let b = l la . Equality if and only if a = I .
3. Add together a2 + b2 :2:: 2ab, b2 + c2 :2:: 2bc, c2 + a2 :2:: 2ca , then
divide by 2.
4. Equivalent to a2b2(a - b)2 :2:: 0 .
5. Multiply a2 + b2 :2:: 2ab by c, b2 + c2 :2:: 2bc by a, c2 + a2 :2:: 2ac by
b, then add.
6. Intermediate step:
(a2 - b2)2 - (a - b)4 = 4ab(a - b)2 .
7. Intermediate step:
(a3 + b3) - (a2b + ab2) = (a + b)(a - b)2 .
8. (3) a = b = c ; (4) a = b or at least one of them = 0 ;
(5) a = b = c ; (6) a = b or at least one of them = 0 ; (7) a = ±b .
ANSWERS TO EXERCI S E S
1 17
Page 24
I. Equivalent to (a - b)2 � 0 . Sign of equality if and only if a = b � 0 .
2. By Theorem 2.7 and the fact that c and d are of the same sign, ac - b<
and ad bel are of the same sign. Multiply out. Sign of equality if and
only if a = b .
3. a2 + b2 � 2ab, a + b � 2 y'ali.
-
4. The equivalence of these inequalities can be proved by multiplying the
first by bd > 0 and the second by I Ibd > 0 ; see Theorem 2.3. The
equivalence of the corresponding equations is proved in the same way.
5. Add I = I to the inequality. Equality alb = cld is equivalent to
ad = bc .
6. Apply Theorem 2. 7 with exponent - I , add I = I to resulting inequality,
then apply Theorem 2.7 again with exponent - I . Sign of equality as in
Exercise 5.
7. Intermediate steps:
a + c _ g_ _ bc - ad > O
b + d b - b(b + d) - '
£_ _ a + c _ be - ad > O .
d b + d - d(b + d) -
8. 30 < 33 ; 22 < 25 ; 1 8 < 2 1 ; 42 < 45 .
9. If a < b and b < c, then a < c .
If a < b and c < d, then a + c < b + d. I f a < b and c is any real
number, then a + c < b + c .
If a < b and c > 0 , then ac < be . If a < b and c < 0 , then
ac > be .
If a < b and c < d, then a - d < b - c . If a < b and c is any real
number, then a - c < b - c .
If 0 < a < b and 0 < c < d, then ac < bd .
If 0 < a < b and 0 < c < d, then aid < blc . In particular, for
a = b = I , if 0 < c < d, then I Id < I I c .
If 0 < a < b, if m and n are positive integers, and if all" and bl ln
denote positive nth roots, then
and
amln < bmln
Chapter 3
Pages 28-29
I. (a) - I (b) 'IT (c) 0 (d) 4 (e) 3 (f) 0 (g) 'IT (h) 0 (i) 4 (j) 3
2. (a) - 7 (b) Vf (c) - 7 (d) O (e) - 3 (f) - 7 (g) O (h) - 7 (i) O (j) - 3
3 . ( - 1)(0) - (1)(0) = 0 .
4. Consider representative cases along with the definition of max {
}.
A N I NT R0 D U CTI 0N T0 I N EQ U ALlT I E S
1 18
5. {a, b, c, d } = { - 1 , - 1 , - 1 , - 1 } .
6. One of {a, b } + and { c, d } + is equal to {a, b, c, d } + ; the other is
nonnegative.
7. First and third inequalities: possibly an additional candidate, namely 0.
Middle inequality: defirution of max {
} and min {
} . No;
a 1. a2, . . . , an would all have to be negative for first strict inequality, and
all positive for third.
8. Multiply inequalities a :2:: b, a :2:: c by - 1 and apply Theorem 2.3.
9. { - a, - b ) - = min {0, - a, - b ) = - max {0, a, b} = - { a, b ) + .
10. Consider representative cases along with definition of max {
}.
Pages 33-34
I. For each real number a, - I a I � - a � I a I · The first sign of equality
holds if and only if a :2:: 0 , and the second if and only if a � 0 .
2.
y
y
3
2
2
X
3
-3
-2
-I
-I
0
I
2
3
-2
( b)
(a)
-3
y
3
2
-3
-2
-1
0
-I
-2
(c)
-3
3. (a) is also the graph of (d), (g); (b) is also the graph of (e), (h); (c) is also
the graph of (f), (i).
y
4
y
3
X
-3
-4
(a)
1 19
A N S W E RS TO E X E RC I S ES
"
"
-5
""
"'
2
/
""
"
( b)
-
y
3
X
(c)
5.
2
"'
-5
( -3 , 4 )
"'
3
X
5
2
-4
/
-4
y
-3
/
/
""
"
/
/
(d)
y
( 3 , 4)
6. Fig. 3.3, odd; Fig. 3.4, odd; Fig. 3.5, even; Fig. 3.6, even.
/
/
X
AN I NTRO D U CTION TO I N EQ U A L I T I ES
1 20
Page 36
I.
y
3
2
-3
0
-I
-2
2
X
3
-I
-2
-3
(a)
'I
y
3
( 3,2)
( - 3, - 2 )
(b)
-3
2.
(- 1 , 1 )
�
y
1,1)
- )
I
I
(e)
(-1,-I
(0, - 1 )
Part (d) i s easy; there is nothing to do.
I
I
( 1 ,0)
(f)J
- .. ( 1 , - 1 )
121
ANSWERS TO EXERCI S E S
3.
y
y
4
a 3
2
-3
(- 1 , - I )
-2
(a)
2
-2
-3
X
-2
-3
( b)
4.
3
y
(-3,
-3
2
I)
-2
-I
-I
0
I
2
3
X
(3, - I )
-2
Page 39
I.
(a) x � O . y 2 0 , x + y � 1 :
(a)
+-----"1�----.... X
-----
(g)
(b) X � 0 , J 2 0 , X + J 2 1 ;
(C )
X 2 0, J 2 0, X +J 2 1 :
(d) x 2 0 , y 2 0 , x + y � 1 ,
(e) x � O , y � O , x + y � 1 ;
(f ) X 2 0 , J � 0 , X + J � 1 ;
(g) X 2 0 , J � 0 , X + J 2 1 .
122
2.
AN INTRODUCTION TO INEQUALITIES
y
y
, 1 2)
(0 /
X
0
-I
·I
-I
(a)
( b)
3.
y
2,5)
5
4
3
l-2, I )
2
4.
(3,5)
5
4
X
y
\
\
\ ! 1,3)
......
....... -......
/
-3
l- 3 ,-4)
-4
/
.1(,!_3. 2)
/
�2,1)
Incomplete graph.
( 0 , - 1 /2 }
X
ANSWERS TO EX ERCI S ES
123
5.
y
6
5
/ -2
Incomplete graph.
Page 4 1
I.
y (0,5)
1/
,."'
,...
"'r--
"'
�fl.
II
I
1_,11
(-5 ,0)
�..��..�
5 1..1
�..��
1\
�
(0,0)
(5,0)
\
',
�
j....
(0, -5)
�..-""'
I.J
IJ
v
X
A N I N TRODU CTION TO I N EQ U A LITI ES
124
2.
y
= 0 . Geometrically, the problem is to determine the locus
of points equidistant from ( - 1 , 0) and ( 1 , 0). Analytically, it is to solve
the equation
3. The y-axis, x
y(x + 1)2 + y2
=
y(x - 1 )2 + y2 .
Page 45
I . (a)
(b)
(c)
2.
3.
4.
5.
= lal
1 -al = � = P
l ab l = y'(aD}2 = P VF£ = l a l · l h l
=
'*' =
=
j(ff {!r
(a) = , (b) > , (c) = ,
(a) > , (b) = , (c) = ,
(a) > , (b) = (c) = ,
(a) = , (b) > , (c) > ,
,
(d)
(d)
=,
=,
f*t·
(e) > .
(e)
(d) > . (e)
(d)
=
,
(e)
=.
=.
=.
� V( ..jQ2 - ..jb2)2 ,
y(a - b)2
� ( ya'£- VJ.i2)2 ,
(a - b)2
a2 - 2ab + b2 � a2 - 2 yla'Illl + b2 ,
2 v'Q'2E'i
� 2ab,
� ab
l ab l
7. Suppose a2 s b2 and ab � 0 . Then
min {a2, b2} = a2 = l a l · l a l S l a l · l b l = ab .
8. If a � 0 , then VQ2 = a; if a < 0 , then yci2 = a > 0 . Sirmlarly,
yci2 is 0 if a = 0 , and otherwise it is the positive member of the set
{a, - a} ; VQ2 = max {a, - a} ; VQ2 = {a, -a}+ ; the graph of
y = yx'I is shown in Fig. 3 6; and VQ2 = a sgn a .
6.
-
A N SWERS TO EXERCI SES
1 25
Chapter 4
Page 49
I . (a) 4, 5; (b) 6, 7.5; (c) 6, 6.5; (d) 0, 10.
2. (a) 3p, 5p; (b) O, p/2; (c) 2p, p2 + I .
Pages 5 1 -52
I . a + b = diameter = 2r, so r = (a + b)/2 ; by similar tnangles, alh =
hlb , so h = vfab.
2. To show that the harmonic mean is less than or equal to the geometric
mean, an intermediate step is
ab(a - b)2 ;:::: 0 ;
or see Exercise I on page 2 1 . The sign of equality holds if and only if
a = b . To show that the harmonic mean is less than or equal to the anth­
metic mean, you can now use the arithmetic-mean-geometric-mean
inequality or can proceed directly, with intermediate step
(a - b)2 ;:::: o .
3. (a) 3.2, 4, 5 ; (b) 4.8, 6, 7.5 ; (c) 5.54-, 6, 6.5; (d) 5.83+, 5.91 +, 6; (e) 6, 6, 6.
4. Half the distance at each rate:
Half the time at each rate:
d = rt = r1t + !JL ,
2
2
+ r2
r = r1
.
2
--
By the result of Exercise 2, half the time at each rate would get you there
sooner.
5. Let b = ! Ia .
Pages 6 1-62
I. In the arithmetic-mean-geometric-mean inequality (4. 1 9) set the first m1
of the numbers a; equal to the same value Yt . set the next m 2 of the num­
bers a1 equal to the same number y2, and set the last mk of the numbers
a1 equal to the same value yk, and observe that
+ mk = n .
m1 + m 2 +
• • •
This gives the first inequality. For the second inequality, set
A N I NTR0DU CTI 0N T0 I N EQ UALITI ES
126
2. 8.5, 9. 1 +; 0.5, 0.7+; p, p.
3. Intermediate step: (a - b)2 � 0 ; or see Exercise l on page 24. Equality
holds if and only if a = b . Since the root-mean-square is greater than
or equal to the arithmetic mean, it is also greater than or equal to the
geometric and harmonic means.
4. (a) A diagonal divides GH into two segments, one of length a/2, the other
of length b/2.
(b ) If ABLK - KLDC, then ABIKL = KLI CD and KL2 = yao.
(c) The harmonic mean is
2ab = h
a+b
To show that
EF = h ,
prove that
·
EO = OF
and use similar triangles:
EO = AE = AC - EC = l _ EC
CD
AC
AC
AC '
But
Hence
EO =
CD
thus
l
_ EO
Eo
or
AB
( A�
+
CD = __!!l!_ = lh
AB + CD a + b 2
EO = A B
and
�)= 1 ;
c
·
EF = 2EO = h .
(d) Set MN = r, let x and y be the altitudes of the constituent trape­
zoids so that x + y is the altitude of ABDC. Then, by hypothesis,
r + a . x _ ! a + b (x
2
- 2
2
+
y)
r + b · y _ ! a + b (x
2
-2
2
'
+
y) .
This system of simultaneous linear equations in x and y has a solution if
and only if
r2 = a2
+
2
b2 .,
hence r is the root-mean-square of a and b.
127
ANSWERS TO EXERCISES
Chapter 5
Page 87
I . Add together ab + be + ca s a2 + b2 + c2 , from Exercise 3 on page
22, and 2(ab + be + ca) = 2(ab + be + ca) , to get 3(ab + be + ca) S
(a + b + c)2 whence
A
,
= 2(ab + be + ca) s
� (a + b + c)2 = �(�Y = �:.
Equality in Exercise 3 on page 22 if and only if a = b
=
c.
2. If the surface area A o f a rectangular box is a fixed value A, then the sum
of the lengths of the twelve edges is at least 2 y'6A, and the box is a
cube if and only if E = 2 y'6A . The proof is immediate from the in­
equality A s £ 2 124 of Exercise I .
3. Let A = area, P = perimeter, a and b = lengths of sides, c = length of
diagonal. By inequality between arithmetic mean and root-mean-square
(Exercise 3 on page 62),
) \ ;'Q2+b2 = 2 Vfc .
; �-2-
(
a+b
P = 2(a + b) = 4 -- 5
2
By inequality between geometric mean and arithmetic mean,
Hence
A
( 1 by = [;.
= ab S a
A
<
c2 .
- 2
The signs of equality hold if and only if a = b . Hence the square has
the greatest perimeter (2 �c) and the greatest area (c2/2) .
Pages 89-90
I . If y = z , then s - y = s - z = x/2 , so
] 2 = s(s - x)(s - y)(s - z) = s(s - x)
� � = � x2(s - x) .
If x = y = z , then s - x = s - y = s - z = s/3 , so
£2 = s(s - x)(s - yXs - z) =
s4
27
.
2. Substitute for £2, J 2, and A 2 from Exercise I , then multiply out and com­
pare terms.
3. Each factor on the right is nonnegative, either because it is a square or
because of its geometric meaning. Have E 2 = I 2 if and only if x = 2s/3,
i.e., if and only if the isosceles triangle actually is equilateral. Have
A 2 if and only if either x = s or y = z , i.e., if and only if the
general triangle actually is isosceles.
J2 =
128
A N I NT R0 D U C TI 0N T0 IN EQ U AL lT l ES
4. The second inequality. See the discussion of / 2 = A2 in the above
solution of Exercise 3.
5. The first inequality. See the discussion of E 2 = I 2 in the above solu­
tion of Exercise 3.
6
.
The formula shows that the square of the area of the triangle is equal to
the square of the area of the equilateral triangle having the same perime­
ter, diminished by a certain amount if the triangle is only isosceles, and
diminished still more if the triangle is not even isosceles.
c = length of hypotenuse, t = length of
altitude. Then by similar triangles, t = able . By the anthmetic-mean­
geometnc-mean inequality,
7. Let a and b = lengths of sides,
1
ab < a2 + b2 £
- c - 2c - 2 ,
with equality if and only if a = b , i.e., if and only if the right tnangle
_
_
is isosceles.
Pages 97-98
I. (5, - 3) .
2 -+- 3 . /iiZ .
. - s y -.v ,
-X + l
3. w
6 - l .
4. Ify is expressed in terms of x from the linear equation, and the resulting
expression is substituted in the quadratic equation, then its discriminant
is 4a2b2n2(a2m 2 + b2n2 - k2) ; this expression vanishes if
k=
±
ya2n(l + b2n2 .
The corresponding double roots are given on page 96, eqs. 5.23.
Chapter 6
Page l l l
I.
Q
/
<
"'
/'
""-/
The city-block length of the diameter is 2 and that of each side is
d1(P1, P2) = d1(0, Q) = 2 . Hence the city-block length of the perimeter
1s
A N S W E R S TO E X E RC I S E S
129
(4)(2) = 8 , and the desired ratio is
r
= 8 = 4.
2
2.
The non-Euclidean length of the diameter is 2 and that of each side is
dx(P1P2) = d'X>(OQ) = 2 . Thus, the non-Euclidean length of the perime­
ter is 8 and
r
3.
Hint.
= 8 = 4.
2
Each regular polygon with an even number of sides and with
center at the origin can be placed in two different positions, each sym­
metric with respect to each coordinate axis. Try to prove that
(a) if the number N of sides of the polygon is divisible by 4, then the
non-Euclidean length of each side is 2 tan ( 1 80° IN ) [and hence the
perimeter has non-Euclidean length 2N tan ( 1 80° IN)] ,
(b) if the number N of sides is even but not divisible by 4, then the
length of each side is 2 sin ( 1 80° I N ) [and hence the perimeter is
2N sin ( 1 80° IN)] ,
(c) the results (a) and (b) are valid for all possible positions of the
polygon.
Since N = 8 is divisible by 4, we have
diameter = 2 ,
non-Euclidean length of perimeter = 1 6 tan ( 1 80° 18)
= 16 tan 22.5° ,
r
16
tan 22.SO
=
= 8( y7
I)
_
2
:::::: 3.3 14 .
4. Since N = 10 is not divisible by 4, we have
diameter = 2 ,
non-Euclidean length of perimeter = 20 sin ( 1 80° I 10) ,
r
= 20 si l 8 o = 10 sin 1 8 o :::::: 3.090 .
f
I ndex
absolute value, 25-45
algebraic characterization of,
city-block distance, I 00-102
classical inequalities, 73-75
complex numbers, 4 1
convex set, 107
cosine inequality, 65
40-
41
and classical inequalities, 73-75
definition of, 26
graph of, 30, 3 1
and sign function, 34-36
addition of inequalities, 1 7
algebraic characterization o fabsolute
value, 40-41
anthmetic mean, 48
anthmetic-geometric mean, 76-78
arithmetic-mean-geometric-mean
inequality, 48-6 1, 73-74
axiom, 7
Dido, problem of, 80-83
three-dimensional version, 86
directed distance, 35
distance, 99-1 1 1
city-block, 1 00- 102
directed, 35
Euclidean, 99- 100
homogeneity of, 100
non-Euclidean, 100-106
positivity of, 100
rotation invariance of, 100
symmetry of, 99
translation invariance of, 99
division of inequalities, 2 1
dual problem, 83
backward induction, 57-59
bound, upper, 12 (footnote)
Buniakowski, V., 63 (footnote)
Cauchy inequality, 62-67, 73-74
geometnc interpretation of, 63
Cauchy-Lagrange identity, 66
ellipsoid,
131
90
132
A N I N TR0 D U CTI 0N T0 I N E Q U AL ITI ES
Euclidean distance, 99- 100
experimentation, mathematical,
Fermat's principle,
forward induction,
84
55-57
48
Lagrange, J. L., 66
"less than" relationship,
9
magnitude, 25
mathemaucal induction, 19
backward, 57-59
Gauss' mean, 76-78
forward, 55-57
geometric mean, 48
maximization,
79ff
geometry of n dimensions, 1 12
maximum function, 27
Graustein, W. C., 65 (footnote)
mean
"greater than" relationship, 5
arithmetic, 48
arithmetic-geometric, 76-78
harmonic mean, 52
of Gauss, 76-78
Holder inequality, 68, 73-74
geometric, 48
homogeneity, 100
harmonic, 52
incidence, angle of, 85
root-mean-square, 61
induction, mathematical, 1 9
minimization, 79ff
backward, 57-59
minimum function, 28
forward, 55-57
Minkowski inequality, 72, 73-74
inequality, 5-6, 9- 10
mixed inequality, 9
addition of, 17
motion, rigid-body, 106
arithmetic-mean-geometric-mean, multiplication of inequalities, 19
48-6 1 , 73-74
multiplication of inequality by a
Buniakowski, 63 (footnote)
number, 1 8
Cauchy, 62-67, 73-74
Cauchy-Schwarz, 63 (footnote)
negation o f inequality, 1 0
division of, 2 1
negative number, 7
graph of, 36-40
product involving a, 10
Holder, 67, 73-74
negative of a number, 7
Minkowski, 72 , 73-74
Niven, Ivan, 12, 6 1
mixed, 9
non-Euclidean distance, 100-106
multiplication of, 19
number
multiplication of, b y a number, 1 8
negative, 10
negation of, 10
negative of a, 10
for powers, 22
negative, product involving a, 10
for roots, 22
pairing of, 7-8
Schwarz, 63 (footnote)
positive, 6
for squares, I I
reciprocal of a, 22
strict, 9
subtraction of, 1 8
order, 12 (footnote)
transitivity law for, 1 6
complete, 1 2 (footnote)
triangle, 42-44, 69-7 1 , 73-74
ordinate, 35
Osgood, W. F , 65 (footnote)
Kazarinoff, N . D., 8 1 , 83
INDEX
pairing of numbers, 7-8
perimeter, 87-88
point set, 106
positive number, 6
powers, 22
product, 10- 1 1
involving a negative number, 10
Pythagorean relationship, 40, 41
ray of light, 83-86
reciprocal, 22
reflection, angle of, 85
refraction, Snell's law of, 86
reverse problem, 83
rigid-body motion, 106
rise, 35
root-mean-square, 6 1
roots, 22
rotation invariance, 100
run, 35
Schwarz, H. A., 63 (footnote)
set
convex, 107
symmetric, 107
sign function, 35-36
133
slope, 35
average, 35
left-hand, 35
right-hand, 35
Snell's law of refraction, 86
specialization, 58
squares, inequality for, I I
square roots, 40-4 1
strict inequality, 9
subtraction of inequalities, 1 8
symmetric means, 7 5
symmetric point set, 107
symmetry, 99
tangent, 93-97
transitivity, 1 6
translation invariance, 99
triangle inequality, 42-44, 69-7 1 ,
73-74
unit circle, 102-106
exterior, 106
interior, 106
unit disc, 106
boundary of, 106
upper bound, 1 2 (footnote)
zero,
6